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>
 
char a[51][51];
char b[51][51];
 
int main(void){
	int n, m;
	int i, j, k, l, f = 0;
	
	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]);
	}
	
	/*printf("a\n");
	for (i = 0; i < n; i++){
		printf("%s\n", a[i]);
	}
	
	printf("b\n");
	for (i = 0; i < m; i++){
		printf("%s\n", b[i]);
	}*/
	
	for (i = 0; i <= n-m; i++){
		for (j = 0; j <= n-m; j++){
			if (a[i][j] == b[i][j]){
				f = 1;
				//printf("f\n");
				for (k = 1; k < m; k++){
					for (l = 1; l < m; l++){
						if (a[i+k][j+l] != b[i+k][j+l]){
							f = 0;
							//printf("%d %d\n", i+k, j+l);
							break;
						}
					}
					if (f == 0) break;
				}
			}
			if (f == 1) break;
		}
		if (f == 1) break;
	}
	
	if (f) printf("Yes\n");
	else printf("No\n");
	
	return 0;
}