Problem:
Score : 200 points

Problem Statement

You are given an image A composed of N rows and N columns of pixels, and a template image B composed of M rows and M columns of pixels.

A pixel is the smallest element of an image, and in this problem it is a square of size 1×1.

Also, the given images are binary images, and the color of each pixel is either white or black.

In the input, every pixel is represented by a character: . corresponds to a white pixel, and # corresponds to a black pixel.

The image A is given as N strings A_1,...,A_N.

The j-th character in the string A_i corresponds to the pixel at the i-th row and j-th column of the image A (1≦i,j≦N).

Similarly, the template image B is given as M strings B_1,...,B_M.

The j-th character in the string B_i corresponds to the pixel at the i-th row and j-th column of the template image B (1≦i,j≦M).

Determine whether the template image B is contained in the image A when only parallel shifts can be applied to the images.

Constraints

1≦M≦N≦50

A_i is a string of length N consisting of # and ..

B_i is a string of length M consisting of # and ..

Input

The input is given from Standard Input in the following format:

N M
A_1
A_2
:
A_N
B_1
B_2
:
B_M

Output

Print Yes if the template image B is contained in the image A. Print No otherwise.

Sample Input 1

3 2
#.#
.#.
#.#
#.
.#

Sample Output 1

Yes

The template image B is identical to the upper-left 2 × 2 subimage and the lower-right 2 × 2 subimage of A. Thus, the output should be Yes.

Sample Input 2

4 1
....
....
....
....
#

Sample Output 2

No

The template image B, composed of a black pixel, is not contained in the image A composed of white pixels.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{
    int n, m, i, j, k, l, count;
    char a[51][51], b[51][51];
    scanf("%d%d", &n, &m);
    for (i=0; i<n; i++)
    {
        scanf("%s", a[i]);
    }
    for (i=0; i<m; i++)
    {
        scanf("%s", b[i]);
    }
    for (i=0; i<=n-m; i++)
    {
        for (j=0; j<=n-m; j++)
        {
            count = 0;
            for (k=0; k<m; k++)
            {
                for (l=0; l<m; l++)
                {
                    if (b[k][l] == a[k+i][l+j])
                    count ++;

                }
            }
            if (count == m*m) {
                break;
            }
        }
    }
    if (count == m*m) {
        printf("Yes\n");
    } else {
        printf("No\n");
    }
    return 0;
}