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

#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a, 0, sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define SIZE(array) (sizeof(array) / sizeof(array[0]))

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

#define MAX_V 100000

#define MOD 1000000007L
#define INF 100000000000

signed main() {
	int n, m;
	cin >> n >> m;
	string a[n], b[m];
	for (size_t i = 0; i < n; i++) {
		cin >> a[i];
	}
	for (size_t i = 0; i < m; i++) {
		cin >> b[i];
	}
	bool finded = false;
	for (size_t i = 0; i < n - m + 1 && !finded; i++) {
		for (size_t j = 0; j < n - m + 1; j++) {
			bool ok = true;
			for (size_t y = 0; y < m && ok; y++) {
				// cout << b[y] << ":" << a[y + i].substr(j, m) << endl << endl;
				if (b[y] != a[y + i].substr(j, m)) {
					ok = false;
				}
			}
			if (ok) {
				finded = true;
				break;
			}
		}
	}
	cout << (finded ? "Yes" : "No") << endl;
}