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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

void prians(bool answer) {
  if(answer)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}
int n, m;
char a[55][55], b[55][55];

bool solve();

int main() {
  cin >> n >> m;
  for(int i = 0; i < n; ++i)
    for(int j = 0; j < n; ++j) cin >> a[i][j];
  for(int i = 0; i < m; ++i)
    for(int j = 0; j < m; ++j) cin >> b[i][j];
  prians(solve());
  return 0;
}

bool solve() {
  for(int i = 0; i + m - 1 < n; ++i)
    for(int j = 0; j + m - 1 < n; ++j) {
      for(int k = 0; k < m; ++k)
        for(int l = 0; l < m; ++l) {
          if(b[k][l] != a[i + k][j + l]) break;
          if(k == l && k == m - 1) return 1;
        }
    }
  return 0;
}