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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
	public static void main (String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int m = scanner.nextInt();
		String [] a = new String [n];
		String [] b = new String [m];
		for (int i = 0 ; i < n ; i++)  a [i] = scanner.next();
		for (int i = 0 ; i < m ; i++)  b [i] = scanner.next();
		
		for (int i = 0 ; i < n - m +1 ; i++) {
			 for (int j = 0 ; j < n - m +1 ; j++){
			 	int count = 0 ;
			 	for (int k = 0 ; k < m ; k++) {
			 		for (int l = 0 ; l < m ; l++){
			 			if (a[k + j].charAt(l + i) == b[k].charAt(l)){
			 				count++;
			 				return;
			 			}
			 		}
			 	}
			 	if (m * m == count) System.out.println("Yes");
			 }
		}
		System.out.println("No");
	}
}