| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the problem, we need to calculate the dimensions of the output after the first layer of the convolutional neural network (CNN) and then flatten it to get the total size of the output vector for each filter. Given: - Input image size: 32x32 - Number of input channels: 1 - Number of filters: 4 - Filter size: 5x5 - Stride: 2 - Zero-padding width: 1 Step 1: Calculate the output size after convolution. The formula to calculate the output size after convolution is: \[ \text{Output size} = \frac{\text{Input size} + 2 \times \text{Padding} - \text{Filter size}}{\text{Stride}} + 1 \] Substituting the given values: \[ \text{Output size} = \frac{32 + 2 \times 1 - 5}{2} + 1 = \frac{32 + 2 - 5}{2} + 1 = \frac{29}{2} + 1 = 14.5 + 1 = 15.5 \] However, the output size should be an integer, so we take the floor value to get 15. Step 2: Calculate the size of the flattened output vector for each filter. Since the input has 1 color channel, and the output after convolution will have a depth equal to the number of filters (4), we have an output tensor of size 15x15x4. To flatten this tensor, we simply multiply the dimensions: \[ \text{Flattened output size} = 15 \times 15 \times 4 = 900 \] Therefore, the total size of the flattened output vector from each filter is 900. However, this is the total size of the output vector when considering all filters combined. The size of the output vector from each individual filter would be: \[ \text{Output vector size per filter} = 15 \times 15 = 225 \] So, the total size of the flattened output vector from each filter is 225. |
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| deepseekr1 | The total size of the flattened output vector from each filter in the first layer of CNN D2 is **900**. Here's the breakdown: 1. **Input dimensions**: 32x32x1 (height, width, channels). 2. **Convolution parameters**: - 4 filters of size 5x5. - Stride = 2. - Zero-padding = 1. 3. **Output spatial dimensions**: - Formula: \( \text{Output size} = \left\lfloor \frac{\text{Input size} + 2 \times \text{Padding} - \text{Filter size}}{\text{Stride}} \right\rfloor + 1 \). - Height/Width calculation: \[ \frac{32 + 2 \times 1 - 5}{2} + 1 = \frac{29}{2} + 1 = 14.5 + 1 = 15.5 \quad (\text{rounded down to 15}). \] - **Output feature map per filter**: 15x15. 4. **Flattened output per filter**: \( 15 \times 15 = 225 \) elements. 5. **Total size across all 4 filters**: \( 225 \times 4 = 900 \). **Answer**: 900. |
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| gpto1 | To determine the size of the flattened output vector from each filter in the convolutional neural network (CNN) described, we'll follow these steps: **Given:** - **Input image size:** \(32 \times 32\) pixels - **Number of color channels:** 1 - **Filter size (F):** \(5 \times 5\) - **Stride (S):** 2 - **Padding (P):** 1 - **Number of filters:** 4 (though this doesn't affect the size from each filter) **Step 1: Calculate Output Feature Map Size** The formula to calculate the output dimension (\(O\)) of a convolutional layer is: \[ O = \left\lfloor \frac{W - F + 2P}{S} \right\rfloor + 1 \] where: - \(W\) is the input size - \(F\) is the filter size - \(P\) is the padding - \(S\) is the stride - \(\left\lfloor \cdot \right\rfloor\) denotes the floor operation, ensuring we have an integer dimension. Plugging in the values: \[ \begin{align*} O &= \left\lfloor \frac{32 - 5 + 2 \times 1}{2} \right\rfloor + 1 \\ &= \left\lfloor \frac{29}{2} \right\rfloor + 1 \\ &= \left\lfloor 14.5 \right\rfloor + 1 \\ &= 14 + 1 \\ &= 15 \end{align*} \] So, the output feature map from each filter will be \(15 \times 15\) pixels. **Step 2: Calculate the Size of the Flattened Output Vector** Since the feature map is two-dimensional (\(15 \times 15\)), flattening it into a vector involves multiplying the height and width: \[ \text{Flattened vector size} = \text{Height} \times \text{Width} = 15 \times 15 = 225 \] **Answer:** The total size of the flattened output vector from each filter is **225**. |
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| GPT-4o | To determine the size of the flattened output vector from each filter in the first layer of the Convolutional Neural Network (CNN) D2, we need to calculate the dimensions of the output feature map after the convolution operation. Here are the steps to solve the problem: 1. **Understand the input dimensions and parameters:** - Input image size: \( 32 \times 32 \) - Number of color channels: 1 - Filter size: \( 5 \times 5 \) - Number of filters: 4 - Stride: 2 - Zero-padding: 1 2. **Calculate the output dimensions of the convolution operation:** The formula to calculate the output width and height of a convolutional layer is given by: \[ \text{Output size} = \left\lfloor \frac{\text{Input size} - \text{Filter size} + 2 \times \text{Padding}}{\text{Stride}} \right\rfloor + 1 \] Plugging in the values for width and height: \[ \text{Output width} = \left\lfloor \frac{32 - 5 + 2 \times 1}{2} \right\rfloor + 1 = \left\lfloor \frac{32 - 5 + 2}{2} \right\rfloor + 1 = \left\lfloor \frac{29}{2} \right\rfloor + 1 = \left\lfloor 14.5 \right\rfloor + 1 = 14 + 1 = 15 \] Since the input image is square, the output height will be the same as the output width: \[ \text{Output height} = 15 \] 3. **Determine the size of the output feature map:** Each filter produces an output of size \( 15 \times 15 \). Since there are 4 filters, each filter's output is independent of the others when we consider the flattened output vector. 4. **Flatten the output feature map:** Flattening the \( 15 \times 15 \) output results in a vector of length: \[ 15 \times 15 = 225 \] Therefore, the total size of the flattened output vector from each filter is \( 225 \). **Summary:** The total size of the flattened output vector from each filter in the first layer of CNN D2 is \( 225 \). |
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| llama3.1-70B-Instruct | To calculate the total size of the flattened output vector from each filter, we need to calculate the output size of the convolutional layer. Given: * Input image size: 32x32 * Number of colour channels: 1 * Filter size: 5x5 * Stride: 2 * Zero-padding width: 1 First, we calculate the output size of the convolutional layer using the formula: Output size = (Input size - Filter size + 2*Padding) / Stride + 1 For the width dimension: Output width = (32 - 5 + 2*1) / 2 + 1 = 16 For the height dimension: Output height = (32 - 5 + 2*1) / 2 + 1 = 16 So, the output size of the convolutional layer is 16x16. Since there are 4 filters, the total size of the flattened output vector from each filter is: 16x16 = 256 Therefore, the total size of the flattened output vector from each filter is 256. |
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| llama3.1-405B-Instruct | To calculate the total size of the flattened output vector from each filter in the first layer of the Convolutional Neural Network (CNN) D2, we need to follow these steps: 1. **Calculate the output size of the convolutional layer**: The formula to calculate the output size (height or width) after a convolutional layer is given by: \[ \text{Output Size} = \frac{\text{Input Size} + 2 \times \text{Padding} - \text{Filter Size}}{\text{Stride}} + 1 \] Given: - Input Size = 32 (both height and width) - Padding = 1 - Filter Size = 5 - Stride = 2 Plugging the values into the formula: \[ \text{Output Size} = \frac{32 + 2 \times 1 - 5}{2} + 1 = \frac{30}{2} + 1 = 15 + 1 = 16 \] So, the output size from each filter is 16x16. 2. **Calculate the total size of the flattened output vector from each filter**: Since each filter produces an output of size 16x16, and there are 4 filters, we calculate the total size of the flattened output vector as follows: - Output from one filter = 16 * 16 = 256 - Since there are 4 filters, the total output size = 256 * 4 = 1024 Therefore, the total size of the flattened output vector from each filter in the first layer of CNN D2 is 1024. |
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| DeepSeek-Coder-V2 | To solve the problem, we need to determine the size of the output from each filter after the convolution operation and then flatten it. Let's break down the steps: 1. **Input Size**: The input images are of size 32x32 with 1 color channel. 2. **Filter Parameters**: - Number of filters: 4 - Filter size: 5x5 - Stride: 2 - Zero-padding: 1 3. **Output Size Calculation**: The output size of a convolutional layer can be calculated using the formula: \[ \text{Output Size} = \left\lfloor \frac{\text{Input Size} - \text{Filter Size} + 2 \times \text{Padding}}{\text{Stride}} \right\rfloor + 1 \] For the height and width dimensions separately: \[ \text{Output Height} = \left\lfloor \frac{32 - 5 + 2 \times 1}{2} \right\rfloor + 1 = \left\lfloor \frac{32 - 5 + 2}{2} \right\rfloor + 1 = \left\lfloor \frac{29}{2} \right\rfloor + 1 = 14 + 1 = 15 \] \[ \text{Output Width} = \left\lfloor \frac{32 - 5 + 2 \times 1}{2} \right\rfloor + 1 = 15 \] Therefore, the output size from each filter is 15x15. 4. **Flattening the Output**: Each filter produces an output of size 15x15. Flattening this gives: \[ \text{Flattened Size} = 15 \times 15 = 225 \] 5. **Total Size of the Flattened Output Vector**: Since there are 4 filters, the total size of the flattened output vector from all filters is: \[ \text{Total Flattened Size} = 4 \times 225 = 900 \] Thus, the total size of the flattened output vector from each filter is: \[ \boxed{900} \] |
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| Qwen2-72B-Instruct | To solve the problem, we need to calculate the dimensions of the output after the first layer of the convolutional neural network (CNN) and then flatten it to get the total size of the output vector for each filter. Given: - Input image size: 32x32 - Number of input channels: 1 - Number of filters: 4 - Filter size: 5x5 - Stride: 2 - Zero-padding width: 1 Step 1: Calculate the output size after convolution. The formula to calculate the output size after convolution is: \[ \text{Output size} = \frac{\text{Input size} + 2 \times \text{Padding} - \text{Filter size}}{\text{Stride}} + 1 \] Substituting the given values: \[ \text{Output size} = \frac{32 + 2 \times 1 - 5}{2} + 1 = \frac{32 + 2 - 5}{2} + 1 = \frac{29}{2} + 1 = 14.5 + 1 = 15.5 \] However, the output size should be an integer, so we take the floor value to get 15. Step 2: Calculate the size of the flattened output vector for each filter. Since the input has 1 color channel, and the output after convolution will have a depth equal to the number of filters (4), we have an output tensor of size 15x15x4. To flatten this tensor, we simply multiply the dimensions: \[ \text{Flattened output size} = 15 \times 15 \times 4 = 900 \] Therefore, the total size of the flattened output vector from each filter is 900. However, this is the total size of the output vector when considering all filters combined. The size of the output vector from each individual filter would be: \[ \text{Output vector size per filter} = 15 \times 15 = 225 \] So, the total size of the flattened output vector from each filter is 225. |
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| baichuan4 | To calculate the total size of the flattened output vector from each filter, we need to determine the dimensions of the output feature map after applying the convolutional layer with the given parameters. Given: - Input image size: 32x32 with 1 color channel - Number of filters: 4 - Filter size: 5x5 - Stride: 2 - Zero-padding: 1 First, let's calculate the dimensions of the output feature map for one filter. Using the formula to find the output height (O_h) and output width (O_w): O_h = (I_h + 2 * P - F_h) / S + 1 O_w = (I_w + 2 * P - F_w) / S + 1 Where: - I_h and I_w are the input height and width (32 in both cases) - P is the padding (1) - F_h and F_w are the filter height and width (5 in both cases) - S is the stride (2) Let's calculate: O_h = (32 + 2 * 1 - 5) / 2 + 1 = 14 O_w = (32 + 2 * 1 - 5) / 2 + 1 = 14 So, the output feature map for each filter will be 14x14. Now, let's calculate the total size of the flattened output vector from each filter: Total size = O_h * O_w * number of filters Total size = 14 * 14 * 4 = 784 Therefore, the total size of the flattened output vector from each filter would be 784. |
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| doubao | The input image is of size 32x32. With zero-padding of width 1, the padded image size becomes (32 + 2*1) x (32 + 2*1) = 34x34. The filter is of size 5x5. With a stride of 2, the output size after convolution for each dimension is (34 - 5) / 2 + 1 = 15. So the output from each filter is 15x15. To get the total size of the flattened output vector, we multiply the dimensions, so 15 * 15 = 225. |
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| minimax | To calculate the total size of the flattened output vector from each filter in the first layer of the CNN D2, we need to determine the size of the feature maps that result from the convolution operation. Given: - Input image size: 32x32 with 1 color channel - Filter size: 5x5 - Number of filters: 4 - Stride: 2 - Zero-padding: width 1 First, let's calculate the size of the feature map after applying the filters with the given parameters. The formula to calculate the output size (W_out) of a convolutional layer is: W_out = ((W_in - F + 2P) / S) + 1 Where: - W_in is the input width (or height, since the image is square) - F is the filter size - P is the padding - S is the stride Using the values from the problem statement: W_out = ((32 - 5 + 2*1) / 2) + 1 W_out = ((32 - 5 + 2) / 2) + 1 W_out = (29 / 2) + 1 W_out = 14.5 + 1 W_out = 15.5 Since we cannot have a fraction of a pixel, we round down to the nearest whole number, which is 15. This means that the width and height of the feature map will both be 15 pixels. Now, since we have 4 filters, we will have 4 feature maps of size 15x15. To find the total size of the flattened output vector from each filter, we simply multiply the dimensions of one feature map: Total size of flattened output vector from each filter = 15 (width) * 15 (height) * 1 (color channels) Total size of flattened output vector from each filter = 225 * 1 Total size of flattened output vector from each filter = 225 So, the total size of the flattened output vector from each filter is 225. |
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| kimi | To calculate the size of the flattened output vector from each filter in the first layer of the CNN D2, we need to consider the following: 1. **Input Image Size**: 32x32 pixels with 1 color channel. 2. **Filter Size**: 5x5 pixels. 3. **Stride**: 2 pixels. 4. **Zero-padding**: 1 pixel of padding on each side. First, let's calculate the output size of the convolution operation. The formula for calculating the output size (H_out) when using zero-padding is: \[ H_{out} = \frac{H_{in} + 2 \times P - F_H}{S} + 1 \] Where: - \( H_{in} \) is the input height (32 pixels). - \( P \) is the padding (1 pixel). - \( F_H \) is the filter height (5 pixels). - \( S \) is the stride (2 pixels). Plugging in the values: \[ H_{out} = \frac{32 + 2 \times 1 - 5}{2} + 1 = \frac{32 + 2 - 5}{2} + 1 = \frac{29}{2} + 1 = 14.5 + 1 = 15.5 \] Since we can't have a fraction of a pixel, the output height will be rounded down to the nearest whole number, which is 15 pixels. Now, let's do the same for the width: \[ W_{out} = \frac{W_{in} + 2 \times P - F_W}{S} + 1 \] Where: - \( W_{in} \) is the input width (32 pixels). - \( F_W \) is the filter width (5 pixels). \[ W_{out} = \frac{32 + 2 \times 1 - 5}{2} + 1 = \frac{32 + 2 - 5}{2} + 1 = \frac{29}{2} + 1 = 14.5 + 1 = 15.5 \] Again, rounding down to the nearest whole number, the output width will be 15 pixels. Now that we have the output size for each filter, we can calculate the total size of the flattened output vector for each filter: \[ \text{Flattened Output Size} = H_{out} \times W_{out} \] \[ \text{Flattened Output Size} = 15 \times 15 = 225 \] So, the total size of the flattened output vector from each filter is 225. Since there are 4 filters, the total size of the output from the first layer (before flattening) would be: \[ \text{Total Output Size} = 4 \times \text{Flattened Output Size} = 4 \times 225 = 900 \] This means that after the first layer, the CNN D2 would have a feature map of size 15x15x4, which when flattened, would result in a vector of size 900. |
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| wenxinyiyan | To calculate the output size of a convolutional layer, we can use the following formula: \[ W_{out} = \frac{W_{in} - F + 2P}{S} + 1 \] \[ H_{out} = \frac{H_{in} - F + 2P}{S} + 1 \] Where: - \(W_{out}\) and \(H_{out}\) are the width and height of the output feature map, respectively. - \(W_{in}\) and \(H_{in}\) are the width and height of the input image, respectively. - \(F\) is the size of the filter (assuming it's a square filter). - \(P\) is the zero-padding added to each side of the input image. - \(S\) is the stride length. Given the problem description, we have: - \(W_{in}\) = 32, \(H_{in}\) = 32 (input image size is 32x32) - \(F\) = 5 (filter size is 5x5) - \(P\) = 1 (zero-padding width is 1) - \(S\) = 2 (stride is 2) Plugging these values into the formulas above, we get: \[ W_{out} = \frac{32 - 5 + 2 \times 1}{2} + 1 = \frac{29}{2} + 1 = 14.5 + 1 = 15.5 \] \[ H_{out} = \frac{32 - 5 + 2 \times 1}{2} + 1 = \frac{29}{2} + 1 = 14.5 + 1 = 15.5 \] However, the dimensions of a feature map must be integers. Since we obtained a non-integer value, it means we need to adjust our calculations. In practice, when applying convolution, if the result is not an integer, it is typically rounded down to the nearest integer. This is because the convolution operation cannot produce fractional pixels. Rounding down \(W_{out}\) and \(H_{out}\) to the nearest integer, we get: \(W_{out}\) = 15 \(H_{out}\) = 15 Therefore, the output feature map from each filter will be of size 15x15. Since there are 4 filters, we will have 4 feature maps of this size. To flatten these feature maps into a vector, we multiply the width and height of each feature map and then multiply by the number of filters: Total size of flattened output vector = 4 * (15 * 15) = 4 * 225 = 900 So, the total size of the flattened output vector from each filter would be 900. Note that this represents the concatenated output of all 4 filters, not the output from each individual filter. If we're interested in the output size from a single filter, it would be 15 * 15 = 225. |
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| zhipu | To determine the total size of the flattened output vector from each filter in the first layer of the Convolutional Neural Network (CNN) D2, we need to calculate the spatial dimensions of the output feature maps produced by the convolution operation. Given: - Input image size: 32x32 with 1 color channel - Number of filters: 4 - Filter size: 5x5 - Stride: 2 - Zero-padding: 1 Let's use the formula for the output size of a convolutional layer: \[ \text{Output size} = \frac{\text{Input size} - \text{Filter size} + 2 \times \text{Padding}}{\text{Stride}} + 1 \] For both the width and height, the calculation is as follows: \[ \text{Output width} = \frac{32 - 5 + 2 \times 1}{2} + 1 = \frac{32 - 5 + 2}{2} + 1 = \frac{29}{2} + 1 = 14.5 + 1 = 15 \] \[ \text{Output height} = \frac{32 - 5 + 2 \times 1}{2} + 1 = \frac{32 - 5 + 2}{2} + 1 = \frac{29}{2} + 1 = 14.5 + 1 = 15 \] So, the output dimensions of each feature map are 15x15. Since there are 4 filters, each producing a 15x15 feature map, the total size of the flattened output vector from each filter is: \[ \text{Size of each flattened output vector} = \text{Output width} \times \text{Output height} = 15 \times 15 = 225 \] Therefore, the total size of the flattened output vector from each filter is 225. Note that this is the size for each individual filter's output. If you were to concatenate the outputs of all 4 filters into a single vector, the total size would be: \[ \text{Total size of concatenated output vectors} = 225 \times 4 = 900 \] However, since the question asks for the size of the flattened output vector from each filter, the answer is 225. |
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| gpto1 | To determine the size of the flattened output vector from each filter in the convolutional neural network (CNN) described, we'll follow these steps: **Given:** - **Input image size:** \(32 \times 32\) pixels - **Number of color channels:** 1 - **Filter size (F):** \(5 \times 5\) - **Stride (S):** 2 - **Padding (P):** 1 - **Number of filters:** 4 (though this doesn't affect the size from each filter) **Step 1: Calculate Output Feature Map Size** The formula to calculate the output dimension (\(O\)) of a convolutional layer is: \[ O = \left\lfloor \frac{W - F + 2P}{S} \right\rfloor + 1 \] where: - \(W\) is the input size - \(F\) is the filter size - \(P\) is the padding - \(S\) is the stride - \(\left\lfloor \cdot \right\rfloor\) denotes the floor operation, ensuring we have an integer dimension. Plugging in the values: \[ \begin{align*} O &= \left\lfloor \frac{32 - 5 + 2 \times 1}{2} \right\rfloor + 1 \\ &= \left\lfloor \frac{29}{2} \right\rfloor + 1 \\ &= \left\lfloor 14.5 \right\rfloor + 1 \\ &= 14 + 1 \\ &= 15 \end{align*} \] So, the output feature map from each filter will be \(15 \times 15\) pixels. **Step 2: Calculate the Size of the Flattened Output Vector** Since the feature map is two-dimensional (\(15 \times 15\)), flattening it into a vector involves multiplying the height and width: \[ \text{Flattened vector size} = \text{Height} \times \text{Width} = 15 \times 15 = 225 \] **Answer:** The total size of the flattened output vector from each filter is **225**. |
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| deepseekv3 | To determine the size of the flattened output vector from each filter in the first layer of the CNN D2, we need to calculate the dimensions of the output feature map produced by each filter. Here's the step-by-step process: ### Given: - **Input image size**: 32x32 (height x width) - **Number of color channels**: 1 (grayscale) - **Number of filters**: 4 - **Filter size**: 5x5 (height x width) - **Stride**: 2 - **Zero-padding**: 1 ### Formula to Calculate Output Feature Map Size: The output height \( H_{\text{out}} \) and width \( W_{\text{out}} \) of the feature map can be calculated using the formula: \[ H_{\text{out}} = \left\lfloor \frac{H_{\text{in}} + 2 \times P - F}{S} \right\rfloor + 1 \] \[ W_{\text{out}} = \left\lfloor \frac{W_{\text{in}} + 2 \times P - F}{S} \right\rfloor + 1 \] Where: - \( H_{\text{in}} \) = Input height (32) - \( W_{\text{in}} \) = Input width (32) - \( F \) = Filter size (5) - \( S \) = Stride (2) - \( P \) = Padding (1) ### Step 1: Calculate Output Height (\( H_{\text{out}} \)) \[ H_{\text{out}} = \left\lfloor \frac{32 + 2 \times 1 - 5}{2} \right\rfloor + 1 \] \[ H_{\text{out}} = \left\lfloor \frac{32 + 2 - 5}{2} \right\rfloor + 1 \] \[ H_{\text{out}} = \left\lfloor \frac{29}{2} \right\rfloor + 1 \] \[ H_{\text{out}} = 14 + 1 = 15 \] ### Step 2: Calculate Output Width (\( W_{\text{out}} \)) \[ W_{\text{out}} = \left\lfloor \frac{32 + 2 \times 1 - 5}{2} \right\rfloor + 1 \] \[ W_{\text{out}} = \left\lfloor \frac{32 + 2 - 5}{2} \right\rfloor + 1 \] \[ W_{\text{out}} = \left\lfloor \frac{29}{2} \right\rfloor + 1 \] \[ W_{\text{out}} = 14 + 1 = 15 \] ### Step 3: Determine the Size of the Flattened Output Vector Each filter produces an output feature map of size \( 15 \times 15 \). When flattened, this becomes a vector of size: \[ \text{Flattened Size} = 15 \times 15 = 225 \] ### Conclusion: The total size of the flattened output vector from each filter is **225**. |
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