PROGRAMMING:Morimori map
Morimori recently wanted to make his circle of friends shine, so he decided to write his own software to beautify photos, and named it morimori Meitu. As we all know, beautifying one's face in a group photo without beautifying the face of the person taking the photo is a great way to occupy the top of the circle of friends, so Mori must have this function. The first step in this feature is to select your face. Sensen first calculates the score of the similarity between all the pixels in an image and the surrounding points. The lower the score, the more "like" a pixel on the edge of an outline. Sensen thinks that the lower the sum of the scores of any continuous pixels, the higher the coincidence degree between the curve and the contour edge. In order to select a complete face, Sensen decided to let the user select two pixels a and B on the face, then the line connecting the two points will divide the image into two parts, and then look for a curve from a to B in the two parts, which coincides with the contour to the highest degree, and the user's face can be spelled out. However, after calculating the score matrix, Sensen suddenly finds that he doesn't know how to find these two curves. Can you help Sensen become the little prince in his circle of friends?
In order to solve the problem conveniently, we make the following supplementary explanation:
-The upper left corner of the image is the coordinate origin (0,0). We assume that all pixels are arranged in matrix format, and their coordinates are non negative integers (that is, the horizontal axis is positive to the right, and the vertical axis is positive to the down).
-Ignore the pixels on the straight line connecting a and B (note that it is not a line segment), that is, the pixels are not considered to be on any partition, so the curve can not pass through the pixels.
-The curve is eight connected (i.e. any pixel can be connected to its surrounding eight pixels). However, in order to calculate accurately, when a pixel is connected to diagonally adjacent oblique pixels, the score * * is increased * * and the $$- sqrt {2} $$of the sum of two pixel scores is reduced by one. For example, in the example, the score of the curve passing through two pixels with coordinates of (3,1) and (4,2) should be the sum of the scores of the two pixels (2 + 2), plus the extra (2 + 2) times ($$\ sqrt {2} - 1 $$), which is about 5.66.
###Input format:
Enter two positive integers, $$n $$and $$M $$($$5 / Le n, M / Le 100 $$) in the first row to represent the number of rows and columns of the pixel score matrix.
Next, there are $$n $$rows, and each row has $$M $$non negative integers no more than 1000, which is the score of the pixel.
The last line gives the coordinates of the start and end pixels selected by the user_{ start}, Y_{ Start}) $$and $$(x)_{ end}, Y_{ end})$$。 The four integers are separated by spaces.
###Output format:
Output the score sum of the contour curve found after dividing the image in one line, keeping two decimal places. Pay attention to the score of starting point and ending point, don't double calculate.
###Input example:
```in
6 6
9 0 1 9 9 9
9 9 1 2 2 9
9 9 2 0 2 9
9 9 1 1 2 9
9 9 3 3 1 1
9 9 9 9 9 9
2 1 5 4
```
###Output example:
```out
twenty-seven point zero four
```
answer：If there is no answer, please comment
In order to solve the problem conveniently, we make the following supplementary explanation:
-The upper left corner of the image is the coordinate origin (0,0). We assume that all pixels are arranged in matrix format, and their coordinates are non negative integers (that is, the horizontal axis is positive to the right, and the vertical axis is positive to the down).
-Ignore the pixels on the straight line connecting a and B (note that it is not a line segment), that is, the pixels are not considered to be on any partition, so the curve can not pass through the pixels.
-The curve is eight connected (i.e. any pixel can be connected to its surrounding eight pixels). However, in order to calculate accurately, when a pixel is connected to diagonally adjacent oblique pixels, the score * * is increased * * and the $$- sqrt {2} $$of the sum of two pixel scores is reduced by one. For example, in the example, the score of the curve passing through two pixels with coordinates of (3,1) and (4,2) should be the sum of the scores of the two pixels (2 + 2), plus the extra (2 + 2) times ($$\ sqrt {2} - 1 $$), which is about 5.66.
###Input format:
Enter two positive integers, $$n $$and $$M $$($$5 / Le n, M / Le 100 $$) in the first row to represent the number of rows and columns of the pixel score matrix.
Next, there are $$n $$rows, and each row has $$M $$non negative integers no more than 1000, which is the score of the pixel.
The last line gives the coordinates of the start and end pixels selected by the user_{ start}, Y_{ Start}) $$and $$(x)_{ end}, Y_{ end})$$。 The four integers are separated by spaces.
###Output format:
Output the score sum of the contour curve found after dividing the image in one line, keeping two decimal places. Pay attention to the score of starting point and ending point, don't double calculate.
###Input example:
```in
6 6
9 0 1 9 9 9
9 9 1 2 2 9
9 9 2 0 2 9
9 9 1 1 2 9
9 9 3 3 1 1
9 9 9 9 9 9
2 1 5 4
```
###Output example:
```out
twenty-seven point zero four
```
answer：If there is no answer, please comment