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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)


import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int m = sc.nextInt();

		String[] n_array = new String[n];
		for(int i = 0; i < n; i++) {
			n_array[i] = sc.next();
		}

		String[] m_array = new String[m];
		for(int i = 0; i < m; i++) {
			m_array[i] = sc.next();
		}

		String check = m_array[0];

		for(int i = 0; i < n - m + 1; i++) { //mの1行目が入る可能性のあるnの行
			List<Integer> indexlist = new ArrayList<>();
			int index = 0;

			while(n_array[i].indexOf(check, index) != -1) {
				index = n_array[i].indexOf(check, index);
				indexlist.add(index);
				index++;
			}

			if(indexlist.size() > 0) {
				for(int ind : indexlist) {
					for(int j = 1; j < m; j++) {
						if(!n_array[i + j].substring(ind, ind + m).equals(m_array[j])) {
							break;
						}
						
					}
					System.out.println("Yes");
					return;
				}
			}
			System.out.println("No");
		}

	}
}