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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
typedef long long LL;
#define rep(i,n) for(LL i=0; i<(LL)(n); i++)

//qsort
int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}

//比較
int max(int a, int b){
	if (a>b) return a;
	else return b;
}

int min(int a,int b) {
	if(a>b) return b;
	else return a;
}

LL ans( LL n,LL xx){
	if(n == -1){
		return 0;
	}
	return (n/xx)+1;
}

//main関数
int main(){
	int M,N;
	scanf("%d %d",&N,&M);
	char A[N][N];
	char B[M][M];
	rep ( i , N ){
		scanf("%s",A[i]);
	}
	rep ( i , M ) {
		scanf("%s",B[i]);
	}
	
	rep( i , N-M+1 ){
		rep ( j , N-M+1){
			int flag = 0;
			rep( k , M ){
				rep( m , M ){
					 if( A[ i+k ][ j+m ] != B[k][m]) flag = 1;
				}
			}
			if(flag == 0){
				printf("Yes");
				return 0;
			}
		}
	}
	printf("No");
	return 0;
}