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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    char a[51][51], b[51][51];
    for(int i = 0; i < n; i++)scanf("%s", a[i]);
    for(int i = 0; i < m; i++)scanf("%s", b[i]);
    int i, j, ii, jj;
    for(i = 0; i < n; i++)
    {
        for(j = 0; j < n; j++)
        {
            if(a[i][j] == b[0][0] && n-i >= m && n-j >= m)
            {
                for(ii = 0; ii < m; ii++)
                {
                    for(jj = 0; jj < m; jj++)
                    {
                        if(a[i+ii][j+jj] != b[ii][jj])break;
                    }
                    if(jj != m)break;
                }
                if(ii == m)
                {
                    printf("Yes");
                    return 0;
                }
            }
        }
    }
    printf("No");
    return 0;
}