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)

// Ver19.03
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define INF (1 << 29) * 2
#define LLINF 4545454545454545454
#define MOD 1000000007
#define ll long long
#define ull unsigned long long
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
int upll(const void *a, const void *b) { return *(ll *)a < *(ll *)b ? -1 : *(ll *)a > *(ll *)b ? 1 : 0; }
int downll(const void *a, const void *b) { return *(ll *)a < *(ll *)b ? 1 : *(ll *)a > *(ll *)b ? -1 : 0; }
void sortup(ll *a, int n) { qsort(a, n, sizeof(ll), upll); }
void sortdown(ll *a, int n) { qsort(a, n, sizeof(ll), downll); }

char a[55][55], b[55][55];
int main()
{
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++)
  {
    scanf("%s", a[i]);
  }
  for (int i = 0; i < m; i++)
  {
    scanf("%s", b[i]);
  }
  for (int i = 0; i <= n - m; i++)
  {
    for (int j = 0; j <= n - m; j++)
    {
      int flag = 1;
      for (int ii = 0; ii < m; ii++)
      {
        for (int jj = 0; jj < m; jj++)
        {
          if (a[i + ii][j + jj] != b[ii][jj])
          {
            flag = 0;
            break;
          }
        }
      }
      if (flag)
      {
        puts("Yes");
        return 0;
      }
    }
  }
  puts("No");
}