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 <cstdio>
#include <iostream>

using namespace std;
int main() {

  int64_t N, M;

  cin >> N >> M;
  char A[N][N], B[M][M];
  
  for(int64_t i=0;i<N;i++)
    for(int64_t j=0;j<N;j++)
      cin >> A[i][j];

  for(int64_t i=0;i<M;i++)
    for(int64_t j=0;j<M;j++)
      cin >> B[i][j];
  
  for(int64_t i=0;i<N-M+1;i++) {
    for(int64_t j=0;j<N-M+1;j++) {
      bool b = true;
      for(int64_t k=0;k<M;k++) {
	for(int64_t l=0;l<M;l++) {
	  if(A[i+k][j+l] != B[k][l]) {
	    b = false;
	    goto jump;
	  }	    
	}
      }
      if(b) {
	cout << "Yes" << endl;
	return 0;
      }
    jump:
      ;
    }
  }
  cout << "No" << endl;
  
  return 0;
}