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>

int main(){
  int N, M, i, j , k , l, flag;
  char A[50][51], B[50][51];
  
  scanf("%d %d", &N, &M);
  for (i = 0; i < N; i++){
    scanf("%s", A[i]);
  }
  for (i = 0; i < M; i++){
    scanf("%s", B[i]);
  }
  
  for (i = 0; i <= N - M; i++){
    for (j = 0; j <= N - M; j++){
      flag = 1;
      for (k = i; k < i + M; k++){
        for (l = j; l < j + M; l++){
          if(A[k][l] != B[k - i][l - j]){
            printf("\n");
            flag = 0;
            break;
          }
        }
        if(flag == 0){
          break;
        }
      }
      if(k == i + M && l == j + M){
        break;
      }
    }
    if(k == i + M && l == j + M){
      break;
    }
  }
  
  if(k == i + M && l == j + M){
    printf("Yes\n");
  } else {
    printf("No\n");
  }
  
  return 0;
}