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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>

int main()
{
    int m,n,ln=0,flag=0,z=0;
    scanf("%d %d",&m,&n);
    char a[m][52],b[n][52];
    for(int i=0;i<m;i++)
        scanf("%s",a[i]);
    for(int i=0;i<n;i++)
        scanf("%s",b[i]);
    for(int x=0;x<=m-n;x++)
    {int ans=0,k=0;
       begain: while(a[x][ans]==b[z][ans])
        {
            ans++;
            if(b[z][ans]=='\0')
            {
                k=1;z++;
                flag++;
                break;
            }
        }
        if(ans<m-n)
        {
            ans++;
            goto begain;
        }
        if(k==0)
        {flag=0;z=0;}
        if(flag==n)
        {
            ln=1;
            break;
        }

    }
    if(ln==1)
        printf("Yes");
    else
        printf("No");
}