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<cstdio>
#include<cstring>
#include<iostream>
int a,b; bool f;
char s[51][51],t[51][51];
using namespace std;
int main()
{
	scanf("%d%d",&a,&b);
	for(int i=1;i<=a;++i)
	for(int j=1;j<=a;++j)
	cin>>s[i][j];
	for(int i=1;i<=b;++i)
	for(int j=1;j<=b;++j)
	cin>>t[i][j];
	for(int i=1;i<=a-b+1;++i)
	for(int j=1;j<=a-b+1;++j)
	if(s[i][j]==t[1][1])
	{
		f=true;
		for(int i1=i;i1<=i+b-1;++i1)
		{
			for(int j1=j;j1<=j+b-1;++j1)
			if(s[i1][j1]!=t[i1-i+1][j1-j+1])
			{
				f=false;
				break;
			}
			if(!f)break;
		}
		if(f)
		{
			printf("Yes");
			return 0;
		}
	}
	printf("No");
}