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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#define FOR(i,k,n) for (int (i)=(k);(i)<(n);(i)++)
using namespace std;

int main()
{
  int N,M;
  cin >> N >> M;

  vector<string> A,B;
  A.resize(N);
  B.resize(M);

  FOR(i,0,N) cin >> A[i];
  FOR(i,0,M) cin >> B[i];


  bool flagg = false;
  FOR(i,0,N-M) {
    FOR(j,0,N-M) {
      // 同じかどうか判定
      bool flag = true;
      FOR(x,0,M) {
        FOR(y,0,M) {
          if (A[i+x][j+y] != B[x][y]) {
            flag = false;
            break;
          }
        }
        if (!flag) break;
      }
      // 一致した・終了して良い
      if (flag) {
        flagg = true;
        break;
      }
    }
  }

  string msg;
  if (flagg) msg = "Yes";
  else msg = "No";

  cout << msg << endl;
  return 0;
}