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++ (Clang 10.0.0) (judge verdict: Accepted)

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
const int Max = 1e6 + 5;
string lst[Max], ls[Max];

int main()
{
	int n, m;
	cin >> n >> m;
	for (int i = 1; i <= n; i++) cin >> lst[i];
	for (int j = 1; j <= m; j++) cin >> ls[j];
	int f = 0;
	for (int i = 1; i <= n - m+1; i++)
	{
		for (int j = 1; j <= n - m+1; j++)
		{
			int tf = 1;
			for (int k = 0; k <= m-1; k++)
			{
				string str;
				str = lst[i + k].substr(j-1, m);
				if (ls[k + 1] != str)
				{
					tf = 0;
					break;
				}
			}
			if (tf == 1) f = 1;
		}
	}
	if (f)cout << "Yes" << endl;
	else cout << "No" << endl;

}