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: Wrong Answer)

#include <iostream>
#include <vector>
using namespace std;

int main() {
    int n, m; cin >> n >> m;
    char A[n + 1][n + 1], B[m + 1][m + 1];
    for (int i = 0; i < n; i++) {
        cin >> A[i];
    }
    for (int i = 0; i < m ; i++) {
        cin >> B[i];
    }
    
    int flag = 1, ans = 0;
    for (int i = 0; i < n - m + 1; i++) {
        for (int j = 0; j < n - m + 1; j++) {
            int nx = i, ny = j;
            for (int k = 0; k < m; k++) {
                if (flag == 0) break;
                for (int l = 0; l < m; l++) {
                    if (A[k + nx][l + ny] != B[k + nx][l + ny]) flag = 0;
                }
                if (k == m - 1 && flag) ans = 1;
            }
        }
    }
    if (ans) puts("Yes");
    else puts("No");
}