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<stdlib.h>
#include<string.h>

int main(void){
  char a[51][51],b[50][51];
  char *p;
  int N,M;
  int i,j,offset;
  int flg=0;

  scanf("%d%d%*c",&N,&M);
  for(i=0;i<N;i++) scanf("%s%*c",a[i]);
  for(i=0;i<M;i++) scanf("%s%*c",b[i]);

  offset=0;
  for(i=0;i+M-1<N;i++){
    if((p=strstr(a[i]+offset,b[0]))==NULL){
      offset=0;
      continue;
    }
    offset=p-a[i];
    flg=1;
    for(j=1;j<M;j++){
      if(strncmp(a[i+j]+offset,b[j],M)!=0){
	flg=0;
	break;
      }
    }
    if(flg==1) break;
    i--;
    offset++;
  }
  puts(flg==1?"Yes":"No");
  return 0;
}