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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,s,n) for(int i = (s); i < (n); i++)

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(18);
  
  int n, m;
  cin >> n >> m;
  char a[n][n];  rep(i, 0, n) {rep(j, 0, n) cin >> a[i][j];}
  char b[m][m];  rep(i, 0, m) {rep(j, 0, m) cin >> b[i][j];}
  
  for(int i = 0; i+m-1 < n; i++) {
    for(int j = 0; j+m-1 < n; j++) {
      bool flag = true;
      rep(k, 0, m) {
        rep(l, 0, m) {
          if(a[i+k][j+l] != b[k][l]) flag = false;
        }
      }
      if(flag) {
        cout << "Yes" << "\n";
        return 0;
      }
    }
  }
  cout << "No" << "\n";
}