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>
#include <stdlib.h>
int main(int argc, char const *argv[]) {
  int i,j,k,l,n,m;
  char s[50][51]={{'\0'}};
  char tmp[50][51]={{'\0'}};
  int x=0,y=0,not=0,match=0;

  scanf("%d %d",&n,&m );


  for(i=0;i<n;i++) scanf("%s",s[i] );
  for(i=0;i<m;i++) scanf("%s",tmp[i] );

  for(i=0;i < n - m + 1;i++){
    for(j=0;j < n - m + 1;j++){
      if( s[i][j] == tmp[0][0]){
        not = 0;
        x = j; y = i;
        for(k=0,y=i;k < m && not!=1;k++,y++){
          for(l=0,x=j;l < m && not!=1;l++,x++){
            //printf("%d %d %d %d %c %c\n",x,y,k,l,s[y][x],tmp[k][l]);
            //printf("%c\n",s[y][x] );
            //printf("%c\n",tmp[k][l] );
            if(s[y][x] != tmp[k][l]){
              not=1;
              break;
            }
          }
        }
        if(not != 1){
          puts("Yes");
          return 0;
        }
      }
    }
  }

  puts("No");
  return 0;
}