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>

int main(){
	int m,n;
	int i,j,k,l;
	int found,end;
	scanf("%d %d",&n,&m);
	
	char M[m][m];
	char N[n][n];
	
	for(i=0;i<n;i++) scanf("%s",N[i]);
	for(i=0;i<m;i++) scanf("%s",M[i]);

	end = 0;
	for(i=0;i<=n-m&&end==0;i++){
		for(j=0;j<=n-m&&end==0;j++){
			found = 1;
			for(k=0;k<m&&found==1;k++){
				for(l=0;l<m&&found==1;l++){
					//printf("%d,%d(%c),%d,%d(%c)\n",i+k,j+l,N[i+k][j+l],k,l,M[k][l]);
					if(N[i+k][j+l]!=M[k][l]) found = 0;
				}
			}
			if(found == 1){
				end = 1;
				break;	
			}
		}
	}
	
	if(found == 1){
		printf("Yes\n");
	}else{
		printf("No\n");
	}
	return 0;
}