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>
#include<vector>
#include<string>
#include<cmath>
#include<algorithm>
#include <numeric>
#include<map>
#include<unordered_map>
#include <queue>
 
using namespace std;
using ll=long long;
#define rep(i,n) for(ll i=0;i<n;++i)
#define all_map(itr,mp) for(auto itr=mp.begin();itr!=mp.end();itr++)
#define ALL(a) (a).begin(),(a).end()

int main(){
    int n, m;
    cin >> n >> m;
    string a[n], b[m];
    rep(i, n)cin >> a[i];
    rep(i, m)cin >> b[i];
    bool flag = false;
    rep(i, n)if(i+m <= n)rep(j, n)if(j+m <= n)if(a[i][j] == b[0][0]){
        int cnt = 0;
        rep(k, m)rep(l, m)if(b[k][l]==a[i+k][j+l])cnt++;
        if(cnt >= m*m)flag=true;
    }
    if(!flag)cout << "No";
    else cout << "Yes";
}