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

int main() {
  int i, j;
  int n, m;
  int temp;
  char flag = 0;

  char a[52][55];
  char b[52][55];

  scanf("%d %d\n", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%[^\n]%*c", a[i]);
  }
  for (i = 1; i <= m; i++) {
    scanf("%[^\n]%*c", b[i]);
  }

  for (i = 1; i <= n - m + 1; i++) {
    flag = 0;
    temp = (char*)strstr(a[i], b[1]) - a[i];
    for (j = 1; j <= m; j++) {
      // if (j == i) {
      // temp = strstr(a[j + i - 1], b[j])-a[j+i-1];
      // printf("%p\n", strstr(a[j + i - 1], b[j])-a[j+i-1]);
      //} else {
      // printf("%p\n", strstr(a[j + i - 1], b[j])-a[j+i-1]);
      if (temp != (char*)strstr(a[j + i - 1], b[j]) - a[j + i - 1]) {
        flag = 1;
        break;
      }
      if ((char*)strstr(a[j + i - 1], b[j]) == NULL){
        flag = 1;
        break;
      }
    }
    //}

    if (flag == 0) {
      break;
    }
  }

  if (flag == 0) {
    printf("Yes");
  } else {
    printf("No");
  }

  return 0;
}