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 <string.h>

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

  char a[n][n+1], b[m][m+1];
  for (i=0; i<n; i++) {
    char t[n+1];
    scanf("%s", t);
    strcpy(a[i], t);
  }
  for (i=0; i<m; i++) {
    char t[n+1];
    scanf("%s", t);
    strcpy(b[i], t);
  }

  char *p;
  for (k=0; k<=n-m; k++) {
    for (i=0; i<=n-m; i++) {
      int flg=0, index, cnt=0;
      for (j=i; j<i+m; j++) {
        if ((p=strstr(&a[i][k], b[j]))!=NULL && flg==0) {
          index=p-a[i];
          cnt++;
          flg=1;
        } else if ((p=strstr(&a[i][k], b[j]))!=NULL) {
          if (p-a[i]==index) {
            cnt++;
            if (cnt==m) {
              printf("Yes\n");
              return 0;
            }
          } else {
            index=p-a[i];
            cnt=1;
          }
        }
      }
    }
  }
  printf("No\n");
  return 0;
}