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 <algorithm>
#include <string>
#include <vector>
#include <iomanip>
#include <cmath>
#include <cstdlib>
#include <map>
#define ll long long
#define REP(i, x) for(int (i) = 0; (i) < (x); (i) ++)
#define pi 3.1415926535
const int MOD = 1000000007;
using namespace std;

int main() {
	int n, m;
	bool flag = false;
	cin >> n >> m;
	vector<string>a(n), b(m);
	REP(i, n) {
		cin >> a[i];
	}
	REP(i, m) {
		cin >> b[i];
	}
	REP(i, n - m + 1) {
		REP(j, n - m + 1) {
			if (a[i][j] == b[0][0]) {
				REP(k, m) {
					if (a[i+k].substr(j, m) != b[k]) {
						flag = false;
						continue;
					}
					else {
						flag = true;
					}
				}
			}
		}
	}
	if (flag) {
		cout << "Yes" << endl;
	}
	else {
		cout << "No" << endl;
	}
	return 0;
}