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<string>

using namespace std;

int main() {
	int n, m;
	string a[50], b[50],ans="No";
	cin >> n >> m;
	for (int i = 0; i < n; i++) {
		string tmp;
		cin >> tmp;
		a[i] = tmp;
	}
	for (int i = 0; i < m; i++) {
		string tmp;
		cin >> tmp;
		b[i] = tmp;
	}
	for (int x = 0; x <= n - m; x++) {
		for (int y = 0; y <= n - m; y++) {
			for (int i = 0; i < m*m; i++) {
				if (a[y + i / m][x + i%m] != b[i / m][i%m]) {
					break;
				}
				if (i == m*m - 1)ans = "Yes";
			}
		}
	}
	cout << ans << endl;
	return 0;
}