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.io.*;
import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);

//   String nyuA=sc.nextLine();
  // String nyuB=sc.nextLine();
 
     String[] S=sc.nextLine().split(" ");

//int n=Integer.parseInt(nyu);

int N=Integer.parseInt(S[0]);
int M=Integer.parseInt(S[1]);

String[][] A=new String[N][N];
String[][] B=new String[M][M];

for(int i=0;i<N;i++){
A[i]=sc.nextLine().split("");
}
for(int i=0;i<M;i++){
B[i]=sc.nextLine().split("");
}
int count=0;

int flag=0;
for(int i=0;i<(N-M)+1;i++){
re:

for(int j=0;j<(N-M)+1;j++){

for(int k=0;k<M;k++){
for(int l=0;l<M;l++){
if(!A[i+k][j+l].equals(B[k][l]) ){
//System.out.println(i+"-"+j+"-"+k+"-"+l);
continue re;
}else{
//System.out.println(i+"-"+j+"-"+k+"-"+l+"-"+A[i+k][j+l]+"--"+B[k][l]);

}

}}
flag=1;
//System.out.println("sor");
}}

//long H=Long.parseLong(t[0]);
//long W=Long.parseLong(t[1]);
//long C=Long.parseLong(t[2]);


if(flag==0)
System.out.println("No");
else
System.out.println("Yes");


	}
}