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,x,y,c=0,i,j,k,l;
	char a[100][100],b[100][100];
	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+1; i++)
	{
		for(j=0; j<n-m+1; j++)
		{
			if(a[i][j]==b[0][0])
			{
				x=0;
				for(k=i; k<i+m; k++)
				{
					y=0;
					for(l=j; l<j+m; l++)
					{
						if(a[k][l]!=b[x][y])
						break;
						y++;
					}
					if(y<m-1)
					break;
					x++;
				}
				if(x==m)
				c=1;
			}
		}
	}
	if(c>0)
	printf("Yes\n");
	else
	printf("No\n");
	
	return 0;
}