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: Runtime Error)


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

int main(){
	int n,m;
	scanf("%d %d",&n,&m);

	char **A,**B;
	int i,j;
	A = malloc(sizeof(char)*n);
	for(i=0;i<n;i++){
		A[i] = malloc(sizeof(char)*n);
	}

	B = malloc(sizeof(char)*m);
	for(i=0;i<m;i++){
		B[i] = malloc(sizeof(char)*m);
	}

	for(i=0;i<n;i++){
		scanf("%s",A[i]);
	}
	for(i=0;i<m;i++){
		scanf("%s",B[i]);
	}

	if(n<m){
		printf("No\n");
		return 0;
	}

	int ib=0;
	int ib_continue=0;
	int is_yes=0;
	for(i=0;i<n;i++){
		for(j=0;j<m;j++){
			if(!strcmp(&A[i][j],B[ib])){
				ib_continue=1;
				ib++;
				if(ib == (m-1)){
					is_yes=1;
				}
				break;
			}else{
				if(j == (m-1) && ib_continue==1){
					printf("No\n");
					return 0;
				}
			}
		}
	}

	if(is_yes){
		printf("Yes\n");
	}else{
		printf("No\n");
	}

	return 0;
}