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: Accepted)

#include <stdio.h>
#include <string.h>

int main(void)
{
	int i, j, l;
	int count = 0;
	int n, m;
	scanf("%d %d", &n, &m);

	char N[100][100];
	char M[100][100];

	for (i = 0; i < n; i++)
		scanf("%s", N[i]);
	for (i = 0; i < m; i++)
		scanf("%s", M[i]);

	if (n >= m)
		for (i = 0; i < n-m+1; i++)
		{
				for (j = 0; j < n-m+1; j++)
				{
					if (strncmp(N[i]+j, M[0], m) == 0)
					{
						count++;
						for (l = i+1 ; l < m+i; l++)
						{
							if (strncmp(N[l]+j, M[l-i], m) == 0)
								count++;
						}

						if (count == m)
						{
							printf("Yes"); return 0;
						}
						count = 0;
					}
				}
		}

	else if (m >= n)
		for (i = 0; i < m-n+1; i++)
		{
				for (j = 0; j < m-n+1; j++)
				{
					if (strncmp(N[i]+j, M[0], n) == 0)
					{
						count++;
						for (l = i+1 ; l < n+i; l++)
						{
							if (strncmp(N[l]+j, M[l-i], n) == 0)
								count++;
						}

						if (count == n)
						{
							printf("Yes"); return 0;
						}
						count = 0;
					}
				}
		}

	printf("No");

	return 0;
}