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>

int main(void)
{
	int n, m;
	char a[51][51];
	char b[51][51];
	int f, g;
	int i, j, k, l;
	
	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]);
	
	for(i=0; i<=(n-m); i++){
		for(j=0; j<=(n-m); j++){
			f = g = 0;
			for(k=0; k<m; k++){
				for(l=0; l<m; l++){
					if(a[i+k][j+l] != b[k][l]){
						f = 1;
						break;
					}
					g = 1;
				}
				
				if(f || g)
					break;
			}
			if(g)
				break;
		}
		if(g)
			break;
	}
	
	if(g)
		puts("Yes");
	else
		puts("No");
	
	
	return 0;
}