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.*;

public class Main{
	public static void main(String[] args){
		Scanner scanner = new Scanner(System.in);
		
		int large = scanner.nextInt();
		//System.out.println(large);
		int small = scanner.nextInt();
		
		char tableL[][] = new char[large][large];
		char tableS[][] = new char[small][small];
		
		for(int i = 0; i < tableL.length; i++){
			String tmp = scanner.next();
			for(int j = 0; j < tableL[i].length; j++){
				tableL[i][j] = tmp.charAt(j);
			}
		}
		
		
		for(int i = 0; i < tableS.length; i++){
			String tmp = scanner.next();
			for(int j = 0; j < tmp.length(); j++){
				tableS[i][j] = tmp.charAt(j);
			}
		}
		
		
		
		for(int i = 0; i < large - small + 1; i++){
			for(int j = 0; j < large - small + 1; j++){
				// ここまでのループで、小さい正方形の左上の座標を網羅する。
				boolean flag = true;
				
				// 関数化しないので注意
				
				for(int k = 0; k < small; k++){
					for(int l = 0; l < small; l++){
						// ここまでのループで指定された左上の座標からの位置を示す。
						//System.out.println("L = " + tableL[i + k][j + l] + " S = " + tableS[k][l]);
						if(tableL[i + k][j + l] != tableS[k][l]){
							flag = false;
							//System.out.println(">> i = " + i + " j = " + j + " k = " + k + " l = " + l);
							break;
						}
					}
					if(flag == false){
						break;
					}
				}
				if(flag == true){
					System.out.println("Yes");
					return;
				}
				
				// 関数の終了
				
				if(flag == false){
					break;
				}
			}
		}
		
		System.out.println("No");
	}
}