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.

Examples

Input

3 2
#.#
.#.
#.#
#.
.#


Output

Yes


Input

3 2
.#
.#.
.#
.
.#


Output

Yes


Input

4 1
....
....
....
....


Output

No

import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		String[] a = new String[n];
		String[] b = new String[m];
		for(int i = 0; i < n; i++) {
			a[i] = sc.next();
		}
		for(int i =0; i < m; i++) {
			b[i] = sc.next();
		}
		String result = "初期化";
		int idx = 0;
out:		for(int i = 0;i <= n-m; i++) {
			//System.out.println("テスト" + i + " idx:" + idx + " → " + a[i].indexOf(b[0], idx) );
			idx = a[i].indexOf(b[0], idx);
			if(idx == -1) {
				//System.out.println("test");
				idx = 0;
			}else{
				//System.out.println(n + " " + m);
				if(n == m) {
					result = "Yes";
					break out;
				}
				for(int j = 1; j <= m-1; j++) {
					//System.out.println("i-:" + i + " idx:" + idx + " 比較" + a[i + j].indexOf(b[j], idx) );
					if(a[i + j].indexOf(b[j], idx) == idx);
					else if(a[i + j].indexOf(b[j], idx) == -1) {
						idx = 0;
						break;
					}else{
						i--;
						idx++;
						break;
					}
					if(j == m-1) {
						result = "Yes";
						break out;
					}
				}
				
			}
			if(i == n-m) {
				result = "No";
			}
		}
		System.out.println(result);
	}
}
