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 C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define rep(i,n) for(i=0;i<n;i++)
#define PI 3.14159265358979323846264338327950L
int main(){
    int i,n,m;
    scanf("%d%d",&n,&m);
    char a[n][n],b[m][m];
    rep(i,n) scanf("%s",a[i]);
    rep(i,m) scanf("%s",b[i]);
    int addx,addy,x,y,match=1;
    rep(addy,n-m+1){
        rep(addx,n-m+1){
            if(addx>=n-m+1||addy>=n-m+1) continue;
            rep(y,m){
                rep(x,m){
                    if(a[y+addy][x+addx]!=b[y][x]) match=0;
                }
            }
            if(match==1){
                printf("Yes");
                return 0;
            }
        }
    }
    printf("No");
    return 0;
}