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.ArrayList;
import java.util.List;
import java.util.Scanner;
 
public class Main {
	static String[] Lb;
	private static Scanner in;
	public static void main(String[] args) {
		in = new Scanner(System.in);
    int N = in.nextInt(); // the number of cells on the Y axis
    int X = in.nextInt(); // the number of cells on the Y axis
    List<String>  lineA= new ArrayList<>();
    List<String>  lineB= new ArrayList<>();
    in.nextLine();
    int count = 0;
    while(count< N){
        lineA.add(in.nextLine());
    count++;
    }
    count =0;
    while(count< X){
        lineB.add(in.nextLine());
    count++;
    }
      System.out.println(recalTo(lineA,lineB));
}
	static int cal (int N,int X){
		int L = 0;
		L = N + X;
		L += (N-X)/X * 2 -X * 2;
		return L;
	}
	//文字列比較
	public static int recal(String A , String B ,int i ,boolean trig){
		if(B.length() + i > A.length()){
			return -1;
		}
		if(A.substring(i, i + B.length()).equals(B)){
			return i;
		}else{
			if(trig){
				return recal(A,B,++i, trig);
			}else{
				return -1;
			}
		}
	}
	
	public static String recalTo(List<String> A , List<String> B){
		int length = 0;
		int index = 0;
		boolean trig = true;
		int lengthLA = 0;
		//Aの続きから検索を再開したい。
		List<String> AA = A;
		while(true){
			if(AA.isEmpty()){
				break;
			}
			if(AA.get(0).length() == lengthLA){
				AA.remove(0);
			}
			String LAA = AA.get(0).substring(lengthLA);
			//length= 0のときは最初の文字から検索して、length>0のときはその位置の文字列を検索する。
			for(String LA : A){
				if(trig){
					LA = LAA;
				}
				String LB  = B.get(index);
				if((length = recal(LA,LB,length, trig)) != -1){
					index++;
					trig = false;
				}else{
					trig = true;
					length = 0;
					if(index != 0){
						break;
					}
				}
				if(B.size() == index && length !=-1 && !trig){
					return "Yas";
				}
			}
			lengthLA++;
		}
		return "No";
	}
}