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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>

char imgA[50][50] = { 0 };
char imgB[50][50] = { 0 };

int main()
{
	int n, m;
	
	scanf( "%d %d", &n, &m );
	
	int i;
	for ( i = 0; i < n; i++ )
	{
		scanf( "%s", imgA[i] );
	}
	for ( i = 0; i < m; i++ )
	{
		scanf( "%s", imgB[i] );
	}
	
	char* p_adrs;
	int j = 0;
	int eq_flg = 0;
	for ( i = 0; i < n - m+1; i++ )
	{
		p_adrs = strstr( imgA[i], imgB[0] );
		
		if ( NULL != p_adrs )
		{
			j = ( imgA[i] - p_adrs ) / sizeof( char* );
			eq_flg = 1;
			break;
		}
	}
	
	if ( 0 == eq_flg )
	{
		printf("No\n");
		return 0;
	}
	
	int b, bj;
	for ( b = 1; b < m; b++ )
	{
		p_adrs = strstr( imgA[i+b], imgB[b] );
		if ( NULL == p_adrs )
		{
			printf("No\n");
			return 0;
		}
		
		bj = ( imgA[i+b] - p_adrs ) / sizeof( char* );
		
		if ( j != bj )
		{
			printf("No\n");
			return 0;
		}
	}
	
	printf("Yes\n");
	
	return 0;
}