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<iostream>
#include<iomanip>
#include<cmath>
#include<string>
#include<cstring>
#include<vector>
#include<list>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
#define fi first
#define se second
#define mp make_pair
#define rep(i, n) for(int i=0;i<n;++i)
#define rrep(i, n) for(int i=n;i>=0;--i)
const int inf=1e9+7;
const ll mod=1e9+7;
const ll mod1=998244353;
const ll big=1e18;
const double PI=2*asin(1);

int main() {
  int N, M;
  cin>>N>>M;
  vector<string> A(N), B(M);
  for(int i=0;i<N;++i) cin>>A[i];
  for(int i=0;i<M;++i) cin>>B[i];
  bool can = false;
  bool tmpcan;
  for(int i=0;i<N-M;++i) {
    for(int j=0;j<N-M;++j) {
      tmpcan = true;
      for(int k=0;k<M;++k) {
        for(int l=0;l<M;++l) {
          if(A[i+k][j+l]!=B[k][l]) tmpcan = false;
        }
      }
      if(tmpcan) can = true;
    }
  }
  if(can) cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
}