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;

int main()
{
  int n, m;
  cin >> n >> m;
  vector<string> a(n);
  vector<string> b(m);
  int i, j, p, q;
  for(i = 0; i < n; i++)
  {
    cin >> a.at(i);
  }
  for(i = 0; i < m; i++)
  {
    cin >> b.at(i);
  }
  int check = 0;
  int s = 0;
  p = 0;
  q = 0;
  while(q <= n-m)
  {
    for(i = 0; i < m; i++)
    {
      for(j = 0; j < m; j++)
      {
        if(a.at(i+p).at(j+q)==b.at(i).at(j))
        {
          check++;
        }
      }
    }
    if(check == m*m)
    {
      s = 1;
      break;
    }
    else
    {
      check = 0;
      p++;
      if(p > n-m)
      {
        p = 0;
        q++;
      }
    }
  }
  if(s == 1)
  {
    cout << "Yes" << endl;
  }
  else
  {
    cout << "No" << endl;
  }
}