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.BufferedReader;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        
		// 1行、2つ(複数)数字
		String[] ss0 = br.readLine().trim().split(" ", 0);
		int N = Integer.parseInt(ss0[0]);
		int M = Integer.parseInt(ss0[1]);

		
		// 2行目以降文字列を配列に格納
		String[] A = new String[N];
		for(int i = 0; i < N; i++){ A[i] = br.readLine(); }
		String[] B = new String[M];
		for(int i = 0; i < M; i++){ B[i] = br.readLine(); }

					//入力値確認
					for(int i=0; i < N; i++){
						//System.out.println(A[i]);
					}
					for(int i=0; i < M; i++){
						//System.out.println(B[i]);
					}
		
		
		String sa = "";
		String sb = "";
		
		
		int a = N-M+1;
		dp("a=["+(a)+"]");
		for(int j=0; j < a; j++){
			for(int i=0; i < a; i++){
				for(int b=0; b < M; b++){
					sb = B[b];
					//文字列切り出し a文字目からb文字目まで
					sa = A[b+i].substring(j, M+j);
					dp("i=[" + i + "], j=[" + j + "], sa=[" + sa + "], sb=[" + sb + "]");
					if ( sa.equals(sb) ){
						dp("ok, b=["+b+"], M-1=["+(M-1)+"]");
						if (b==(M-1)) { System.out.println("Yes"); return; }
					} else {
						dp("NG");
						if (i==(a-1)) { break; }
						//break;
					}
				}
			}
		}

		System.out.println("No"); return;


    }
    // Debug.Print
    static void dp(String s) {
		//System.out.println(s);
	}
    static void dp(StringBuilder s) { dp(s.toString()); }
    static void dp(int i) { dp(String.valueOf(i)); }
    static void dp(long i) { dp(String.valueOf(i)); }

}