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.io.*;
import java.util.*;
public class Main {
	
	public static void main(String[] args) {
	
		MyScanner sc = new MyScanner();
		
		int p=sc.nextInt();
		int q=sc.nextInt();
		String a[]=new String[p];
		String b[]=new String[q];
		for(int i=0;i<p;i++) {
			a[i]=sc.next();
		}
		for(int i=0;i<q;i++) {
			b[i]=sc.next();
		}
		String ans="No";
		for(int i=0;i<=p-q;i++) {
			for(int j=0;j<=p-q;j++) {
				if(a[i].substring(j,j+q).equals(b[0])) {
					ans="Yes";
					for(int k=i;k<i+q;k++) {
						if(!a[k].substring(j,j+q).equals(b[k-i])) {
							
							ans="No";
						}
					}
				}
			}
		}
		
		
        System.out.println(ans);
		
	}
	
	
	
	
//----------------------------------------------------------
	static int l_min(int[] a) {
		Arrays.sort(a);
		return a[0];
	}
	static int l_max(int[] a) {
		int l=a.length;
		Arrays.sort(a);
		return a[l-1];
	}
	public static PrintWriter out;

	public static class MyScanner {
		BufferedReader br;
		StringTokenizer st;

		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}

	}
	

}