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>
using namespace std;

int N;
int M;
char A[ 60 ][ 60 ];
char B[ 60 ][ 60 ];

bool Check( const int x, const int y )
{
int i;
int j;

	for( i = 1; i < M; i++ )
	{
		for( j = 0; j < M; j++ )
		{
			if( A[ y + i ][ x + j ] != B[ i ][ j ] )
				return false;
		}
	}

	return true;
}

int main( int argc, char *argv[] )
{
int i;
int x;
char *p;

	ios::sync_with_stdio( false );
	cin.tie( 0 );

	cin >> N;
	cin >> M;
	for( i = 0; i < N; i++ ) cin >> A[ i ];
	for( i = 0; i < M; i++ ) cin >> B[ i ];

	for( i = N - M; i >= 0; i-- )
	{
		x = 0;
		p = strstr( A[ i ], B[ 0 ] );
		while( p != NULL )
		{
			x = p - A[ i ];
			if( Check( x, i ) )
			{
				cout << "Yes" << endl;

				return 0;
			}

			x++;
			while( x > N - M ) break;

			p = strstr( A[ i ] + x, B[ 0 ] );
		}
	}

	cout << "No" << endl;

	return 0;
}