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;
    char N[50][50], M[50][50];
    cin>>n>>m;
    for(int i=0; i<n;i++){
        for(int j=0; j<n; j++){
            cin>>N[i][j];
        }
    }
    for(int i=0; i<m;i++){
        for(int j=0; j<m; j++){
            cin>>M[i][j];
        }
    }
    for(int i=0; i<n-m+1; i++){
        for(int j=0; j<n-m+1; j++){
            bool ok=true;
            for(int k=i; k<i+m;k++){
                for(int l=j; l<j+m; l++){
                    if(N[k][l]!=M[k-i][l-j]){ok=false; break;}
                }
            }
            if(ok==true){cout<<"Yes"; return 0;}
        }
    }
    cout<<"No";
    return 0;
}