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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

int main()
{
    int n, m;
    cin >> n >> m;
    char a[n][n], b[m][m];

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            cin >> a[i][j];
        }
    }

    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < m; j++)
        {
            cin >> b[i][j];
        }
    }
    int /*my,mx,*/c=0;
    for (int ny = 0; ny < n; ny++)
    {
        for (int nx = 0; nx < n; nx++)
        {
            for (int my = 0; my < m; my++)
            {
                for (int mx = 0; mx < m; mx++)
                {   
                    if(a[ny+my][nx+mx]==b[my][mx]) c++;
                }
            }
            if(c==m*m) 
            {
                cout << "Yes" << endl;
                return 0;
            }
            /*cout << ">>>" << 'n' << ny+my << ',' << nx+mx << '|' << 'm' << my << ',' << mx << ':' << a[ny+my][nx+mx] << '|' << b[my][mx] << "C=" << c << endl;*/
            c = 0;
        }
    }
    cout << "No" << endl;
    return 0;
}