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>
#include<stdlib.h>

int main(){
	int m,n;
	int i,j,k,l;
	int flag = 0;
	char** a;
	char** b;
	char* str;

	scanf("%d %d",&n,&m);
	a = (char**)calloc(n,sizeof(char*));
	for(i=0; i<n; i++){
		a[i] = (char*)calloc(n,sizeof(char));
	}
	b = (char**)calloc(m,sizeof(char*));
	for(i=0; i<m; i++){
		b[i] = (char*)calloc(m,sizeof(char));
	}

	str = (char*)calloc(n+1,sizeof(char));
	for(i=0; i<n; i++){
		scanf("%s",str);
		for(j=0; j<n; j++){
			a[i][j] = str[j];
		}
	}


	str = (char*)calloc(m+1,sizeof(char));
	for(i=0; i<m; i++){
		scanf("%s",str);
		for(j=0; j<m; j++){
			b[i][j] = str[j];
		}
	}

	i = 0;
	while( i <= n - m && flag == 0 ){
		j = 0;
		while( j <= n - m && flag == 0 ){
			flag = 1;
			k = 0;
			while( k < m && flag == 1 ){
				l = 0;
				while( l < m && a[i+k][j+l] == b[k][l] ){
					l++;
				}

				if( l == m ){
					flag = 1;
				}else{
					flag = 0;
				}

				k++;
			}
			j++;
		i++;
		}
	}

	if( flag == 1 ){
		printf("Yes\n");
	}else{
		printf("No\n");
	}

	return 0;
}