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: Accepted)

#pragma region
#include <bits/stdc++.h>
using namespace std;
inline void ci(void){
  return;
}
template <typename First, typename... Rest>
void ci(First& first, Rest&... rest){
  cin >> first;
  ci(rest...);
  return;
}
inline void co(void){
  cout << "\n";
  return;
}
template <typename First, typename... Rest>
void co(First first, Rest... rest){
  cout << first << " ";
  co(rest...);
  return;
}
#define ll long long
#define ld long double
#define st string
#define dl double
#define wh while
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define tp tuple
#define ft first
#define sc second
#define ve vector
#define tos to_string
#define len length
#define rt return 0
#define br break
#define con continue
#define lcm(n, m) (n * m / gcd(n, m))
#define np next_permutation
#define repi(i, m, n) for(int i=(int)(m);i<(int)(n);i++)
#define rep(i, n) repi(i, 0, n)
#define repill(i, m, n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define repll(i, n) repill(i, 0, n)
#define revi(i, m, n) for(int i=(int)(m);i>(int)(n);i--)
#define rev(i, n) revi(i, 0, n)
#define revill(i, m, n) for(ll i=(ll)(m);i>(ll)(n);i--)
#define revll(i, n) revill(i, 0, n)
#define sz(v) ((int)(v).size()) 
#define each(i, n) for(auto&& i: n)
#define all(n) (n.begin(), n.end())
#define allr(n) (n.rbegin(), n.rend())
#define sp(n) cout << fixed << setprecision(n)
#pragma endregion
// sqrt ルート, int 2*10^9, gcd　最大公約数
// __builtin_popcount bitの個数

int main() {
  int N, M;
  ci(N, M);
  ve<ve<char>> A(N, ve<char>(N)), B(M, ve<char>(M));
  rep(i, N){
    st a;
    ci(a);
    rep(j, N){
      A.at(i).at(j) = a.at(j);
    }
  }
  rep(i, M){
    st b;
    ci(b);
    rep(j, M){
      B.at(i).at(j) = b.at(j);
    }
  }
  rep(i, N - M + 1){
    rep(j, N - M + 1){
      bool f = 1;
      rep(k, M){
        rep(l, M){
          if(A.at(i + k).at(j + l) != B.at(k).at(l)){
            f = 0;
          }
        }
      }
      if(f){
        co("Yes");
        rt;
      }
    }
  }
  co("No");
}