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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>

#define _CRT_SECURE_NO_WARNINGS
#define TLong long long
#define TBMod 1000000007

int main(int argc, char const *argv[]){
	int n,m;
	int a[57][57],b[57][57];
	int f;
	scanf("%d%d",&n,&m);

	for(int i = 0; i<n; ++i){
		scanf("%s",a[i]);
	}
	for(int i = 0; i < m; ++i){
		scanf("%s",b[i]);
	}
	for(int i = 0; i <= n - m; ++i){
		for(int l = 0; l <= n - m; ++l){
			f = 0;
			for(int j = 0; j < m; ++j){
				for(int k = 0; k < m; ++k){
					if(a[j + i][k + l] != b[j][k])
					{
						f = 1;
						break;
					}
				}
			}
			if(f == 0)
			{
				puts("Yes");
				return 0;
			}
		}
	}
	puts("No");
	return 0;
}