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)

//失敗するからこそそこから立ち向かって行く強さがあってそんな強さが本当の強さだと私は思うから
// ぜったいあきらめない
#include<bits/stdc++.h>
using namespace std;
 
#if defined(DAIJOBU)
#include "/home/ichigo/debug.hpp"
#define deb(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define deb(x...)
#endif
#define FIO {ios::sync_with_stdio(0);cin.tie(nullptr);cout.tie(nullptr);}

vector<string> A(55), B(55);

int main(){
	FIO
	
	int n, m;cin >> n >> m;
	for(int i = 0; i < n; i++)cin >> A[i];
	for(int i = 0; i < m; i++)cin >> B[i];
	
	for(int i = 0; i <= n - m; i++){
		for(int j = 0; j <= n - m; j++){
			int c = 0;
			for(int k = i; k < i + m; k++){
				for(int l = j; l < j + m; l++){
					if(A[k][l] == B[k - i][l - j])c++;
				}
			}
			if(c == m * m){
				puts("Yes");
				return 0;
			}
		}
	}
	puts("No");
	return 0;
}