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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
typedef long long ll;
typedef long double ld;
int  main()
{
	int n, m; cin >> n >> m;
	vector<string> a(n);
	vector<string> b(m);
	rep(i, n)
	{
		cin >> a[i];
	}
	rep(i, m)
	{
		cin >> b[i];
	}
	int cnt = 0;
	rep(i, n)
	{
		rep(j, n-m+1)
		{
			int k = 0;
			if (a[i].substr(j, m) == b[k])
			{
				cnt = 0;
				for(k = 0; k < m; k++,i++)
				{
					if (a[i].substr(j, m) == b[k])
						cnt++;
				}
				if (cnt == m)
				{
					cout << "Yes" << endl;
					return 0;
				}
			}
		}
	}
	cout << "No" << endl;
	// cout << cnt << endl;
	return 0;
}