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


int main(){
	int N,M;
	scanf("%d %d",&N,&M);

	char A[50][50];
	char B[50][50];
	char input_A[50];
	char input_B[50];

	for(int i = 0; i < N; i++){
		scanf("%s",input_A);
		for(int j = 0; j < N; j++){
			if(input_A[j] == '#'){
				A[i][j] =  '#';
			}
			else{
				A[i][j] = '.';
			}
		}
	}
	for(int i = 0; i < M; i++){
		scanf("%s",input_B);
		for(int j = 0; j < M; j++){
			if(input_B[j] == '#'){
				B[i][j] =  '#';
			}
			else{
				B[i][j] = '.';
			}
		}
	}

	for (int i = 0; i < N; i++){
		// printf("A[%d] = %s\n",i,A[i]);
	}
	for (int i = 0; i < M; i++){
		// printf("B[%d] = %s\n",i,B[i]);
	}

	int k = 0;
	char *s;
	int place;

	for (int i = 0; i < N; i++){
		s = strstr(A[i],B[k]);
		place = s -A[i];

		/*printf("%d times s = %s\n",i,s);
		printf("place = %d\n",place);
		printf("A[%d] = %s\n",i,A[i]);
		printf("B[%d] = %s\n",k,B[k]);*/
		if(s != NULL){
			k++;
		}
		else{
			k = 0;
		}
		if(k == M){
			printf("Yes\n");
			return 0;
		}
	}

	printf("No\n");

	return 0;
}