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<iostream>

using namespace std;



int main(){

    int N,M,dy=0; cin >> N >> M;
    char A[N][N],B[M][M];
    if(N < M){
      cout <<"No"<<endl;
      return 0;
    }

//画像A,Bの入力
    for(int i=0; i<N; i++){
        cin >> A[i];
    }
    for(int i=0; i<M; i++){
      //for(int j=0; j<N; j++){
        cin >> B[i];
      //}
    }
/*
    for(int k=0; k<N; k++){
      for(int l=0; l<N; l++){
          cout << A[k][l];
      }
         cout << endl;

    }


        for(int k=0; k<M; k++){
          for(int l=0; l<M; l++){
              cout << B[k][l];
          }
             cout << endl;

        }
        */
  for(int i=M; i<=N; i++){
    int dx = 0;
    for(int j=M; j<=N; j++){
      int Tcon = 0;
      //Bに合致するか比較
      for(int k=0; k<M; k++){
        for(int l=0; l<M; l++){

            if(A[k+dx][l+dy] == B[k][l]){
              Tcon += 1;
              /*cout << "Tcon " << Tcon << endl;
              cout << "K " << k << "l"<< l<< endl;
              cout << A[k+dx][l+dy] <<B[k][l] <<endl;*/
              if(Tcon == M*M){
                cout << "Yes" << endl;
                return 0;
              }
            }
        }

      }

      dx +=1;
    }

    dy += 1;
  }

  cout <<"No" << endl;


}