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>

int main(void){
 int N, M;
 int i, j, k, l;
 int flag;
 char **A, **B;
 scanf("%d%d%[^\n]%*c", &N, &M);
 A = (char **)malloc(sizeof(char *)*N);
 for(i = 0; i < N; i++) A[i] = (char *)malloc(N);
 B = (char **)malloc(sizeof(char *)*M);
 for(i = 0; i < M; i++) B[i] = (char *)malloc(M);
 for(i = 0; i < N; i++){
  for(j = 0; j < N; j++) A[i][j] = getchar();
  getchar();
 }
 for(i = 0; i < M; i++){
  for(j = 0; j < M; j++) B[i][j] = getchar();
  getchar();
 }

 for(i = 0; i <= N - M; i++){
  for(j = 0; j <= N - M; j++){
   flag = 1;
   for(k = 0; k < M; k++){
    for(l = 0; l < M; l++){
     if(A[k + i][l + j] != B[k][l]) flag = 0;
    }
   }
   if(flag){
    printf("YES\n");
    return 0;
   }
  }
 }
 printf("NO\n");
 return 0;
}