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;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)

int main(){
    int n,m; cin >> n >> m;
    vector<vector<char>> A(n,vector<char>(n)),B(m,vector<char>(m));
    rep(i,n){
        rep(j,n){
            cin >> A[i][j];
        }
    }
    rep(i,m){
        rep(j,m){
            cin >> B[i][j];
        }
    }
    bool flag = false;
    vector<vector<char>> C(m,vector<char>(m));

    for(int i = 0; i < n-m+1; i++){
        for(int j = 0; j < n-m+1; j++){

            for(int k = 0; k < m; k++){
                for(int l = 0 ; l < m; l++){
                    C[k][l] = A[k+i][l+j];
                }
            }
            flag = true;
            for(int k = 0; k < m; k++){
                for(int l = 0; l < m; l++){
                    if(B[k][l] == C[k][l]){
                        continue;
                    }else{
                        flag = false;
                        break;
                    }
                }
                if(!flag){
                    break;
                }   
            }
            if(flag){
                break;
            }
        }
        if(flag){
            break;
        }
    }

    if(flag){
        cout << "Yes" << endl;
    }else{
        cout << "No" << endl;
    }

    
}