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>
#include<string.h>

char a[51][51];
char b[51][51];

int main(void) {

	int i, j, k, l;
	int n, m;
	int flag = 0;
	char c;

	scanf("%d%d%c", &n, &m, &c);

	//行列のnまたはm成分には改行文字が含まれていることに注意//
	for (i = 0; i < n; i++) {
		scanf("%s", a[i], 51);
	}

	for (i = 0; i < m; i++) {
		scanf("%s", b[i], 51);
	}


	//同じ行列が含まれるか判定//
	//まず大きいほうの各行列成分について//
	for (i = 0; i <= n - m; i++) {
		for (j = 0; j <= n - m; j++) {
			//小さいほうの各行列成分が等しいか確認//
			for (k = 0; k < m; k++) {
				for (l = 0; l < m; l++) {
					if (a[i + k][j + l] != b[k][l]) {
						flag = 1;
						break;
					}
				}
				if (flag == 1) {
					break;
				}
			}
			if (flag == 0) {
				printf("Yes\n");
				return 0;
			}
			flag = 0;
		}
	}

	printf("No\n");
	   
	return 0;
}