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(int argc, char *argv[])
{
	int i, j, k, M, N;
	char mm[64][64], nn[64][64];
	scanf("%d %d\n", &N, &M);
	for ( i = 0; i < N; i++ ){
		fgets(nn[i], 64, stdin);
	}
	for ( i = 0; i < M; i++ ){
		fgets(mm[i], 64, stdin);
	}
						 
	for ( i = 0; i + M <= N; i++ ){
		for ( j = 0; j + M <= N; j++ ){
			for ( k = 0; k < M; k++ ){
				if ( memcmp(nn[k+i]+j, mm[k], M) != 0 ){
					break;
				}
			}
			if ( k == M ){
				printf("Yes\n");
				return 0;
			}
		}
	}
	printf("No\n");
	return 0;
}