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 A_wh,B_wh,i,j,k,l,mas=0,ans=0;
	char mapA[50][50],mapB[50][50];
	
	scanf("%d %d ",&A_wh,&B_wh);
	
	for(i=0;i<A_wh;i++){
		for(j=0;j<=A_wh;j++){
			mapA[i][j]=getchar();
		}
	}
	
	for(i=0;i<B_wh;i++){
		for(j=0;j<=B_wh;j++){
			mapB[i][j]=getchar();
		}
	};
	for(k=0;k<A_wh;k++){
		for(l=0;l<A_wh;l++){
			for(i=0;i<B_wh;i++){
				for(j=0;j<B_wh;j++){
					if(mapA[k+i][l+j]!=mapB[i][j]){
						mas=0;
						break;
					}
					else{
						mas++;
					}
				}
			}
			if(mas==B_wh*B_wh){
				ans=1;
				break;
			}
		}
	}
	if(ans==1){
		printf("Yes\n");
	}
	else{
		printf("No\n");
	}
	
	return 0;
}