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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
	int n,m;
	char a[55][55];
	char b[55][55];
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(int i=1;i<=n;i++)
		scanf("%s",a[i]+1);
		for(int i=1;i<=m;i++)
		scanf("%s",b[i]+1);
		int flag=0;
		for(int e=1;e<=n-m+1;e++)
		for(int i=1;i<=n-m+1;i++)
		{
			int cnt=0;
			for(int j=1;j<=m;j++)
			for(int k=1;k<=m;k++)
			{
				if(b[j][k]==a[e+j-1][i+k-1])
				cnt++;
			}
			if(cnt==m*m)
			flag=1;
		}
		if(flag)
		printf("Yes\n");
		else
		printf("No\n");
	}
	return 0;
 }