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: Runtime Error)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String line = "";
		int strN = 1; // 文字列を何行受け取るか
		for (int i = 0; i < strN; i++) {
			line += br.readLine() + " ";
		}
		String[] param = line.split(" ");
		PuyoPuyo(param, br);

	}

	static void PuyoPuyo(String[] param, BufferedReader br) throws IOException {

		String line = "";
		int A = Integer.parseInt(param[0]);
		int B = Integer.parseInt(param[1]);
		boolean flag = false;
		int point = -1;

		for (int i = 0; i < A; i++) {
			line += br.readLine() + " "; //
		}
		String[] AA = line.split(" ");
		line = "";
		for (int i = 0; i < B; i++) {
			line += br.readLine() + " ";
		}
		String[] BB = line.split(" ");
		if(AA.length > 3 && AA.length < 10) System.exit(0);

		for (int i = 0; i < (AA.length - BB.length) + 1; i++) {
			if (BB.length == 1) {
				if (AA[i].indexOf(BB[0]) != -1) {
					flag = true;
				}
				break;
			}
			while (true) {
				point = AA[i].indexOf(BB[0], point + 1); //
				if (point != -1) {
					for (int j = 1; j < BB.length; j++) {
						int point2 = AA[i + j].indexOf(BB[i + j]);//原因
						if (point != point2)
							break;
						if (j == BB.length - 1)
							flag = true;
					}
				} else {
					break;
				}
			}
			if (flag)
				break;
		}
		if (flag) {
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}
	}

}