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: Accepted)

#include<stdio.h>

int main(){
	int N, M;
	int i, j, k, l, sc;
	_Bool t = 0;
	sc = scanf("%d %d", &N, &M);
	
	char A[N][N + 1], B[M][M + 1];
	
	for(i = 0; i < N; i ++){
		sc = scanf("%s", &A[i][0]);
	}
	for(i = 0; i < M; i ++){
		sc = scanf("%s", &B[i][0]);
	}
	
	for(k = 0; k < N - M + 1; k ++){
		for(l = 0; l < N - M + 1; l ++){
			t = 0;
			for(i = 0; i < M; i ++){
				for(j = 0; j < M; j ++){
					if(B[i][j] != A[k + i][l + j]){
						t = 1;
						break;
					}
				}
				if(t == 1) break;
			}
			if(i == M && j == M){
				printf("Yes\n");
				return 0;
			}
		}
	}
	printf("No\n");	
	return 0;
}