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 <iostream>
#include <algorithm>

using namespace std;

int main(int argc, const char * argv[]) {
  int N,M;
  cin >> N >> M;
  string ori[50];
  string temp[50];
  for(int i=0; i<N; i++){
    cin >> ori[i];
  }
  for(int i=0; i<M; i++){
    cin >> temp[i];
  }

  for(int i=0; i<M; i++){
    for(int j=0; j<N-M; j++){
      if(ori[i].substr(j,M) == temp[i]){
        //一番上の行が一致した時
        bool notMatchFlag = false;
        for(int k=i+1; k<M; k++){
          if(!(ori[k].substr(j,M) == temp[k])){
            notMatchFlag = true;
            break;
          }
        }
        if(notMatchFlag == false){
          cout << "Yes" << endl;
          return 0;
        }
      }
      else continue;
    }
  }
  cout << "No" << endl;
  return 0;

}