Problem: 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

Solution (in JAVA):

Solution:
import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
    
		// 整数の入力
		int n = sc.nextInt();
		int m = sc.nextInt();
        String[] aList = new String[n];
        String[] bList = new String[m];
      
        for (int i = 0; i < n; i++) {
          aList[i] = sc.next();
        }
        for (int i = 0; i < m; i++) {
          bList[i] = sc.next();
        }
        String ans = "No";
        hLoop: for (int h = 0; h < (n - m + 1); h++) {
          wLoop: for (int w = 0; w < (n - m + 1); w++){
             if (!aList[h].substring(w, w + m).equals(bList[0])) {
               continue wLoop;
             }
             int candidate = h;
             for (int index = 0; index < m; index++) {
               if (!aList[candidate].substring(w, w + m).equals(bList[index])) {
                 continue wLoop;
               }
               candidate++;
               if (index == m - 1) {
                 ans = "Yes";
                 break hLoop;
               }
             }
          }
        }
    
		// 出力
		System.out.println(ans);
	}
}