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 <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define rrep(i, n) for(int i = 1; i <= (int)(n); i++)
#define drep(i, n) for(int i = (n)-1; i >= 0; i--)
#define ALL(x) (x).begin(), (x).end()
#define dup(x,y) (((x)+(y)-1)/(y))
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

// Mod int
const int mod = 1000000007;
// const int mod = 998244353;
struct mint {
  ll x;
  mint():x(0){}
  mint(ll x):x((x%mod+mod)%mod){}
  // mint(ll x):x(x){}
  mint& fix() { x = (x%mod+mod)%mod; return *this;}
  mint operator-() const { return mint(0) - *this;}
  mint& operator+=(const mint& a){ if((x+=a.x)>=mod) x-=mod; return *this;}
  mint& operator-=(const mint& a){ if((x+=mod-a.x)>=mod) x-=mod; return *this;}
  mint& operator*=(const mint& a){ (x*=a.x)%=mod; return *this;}
  mint operator+(const mint& a)const{ return mint(*this) += a;}
  mint operator-(const mint& a)const{ return mint(*this) -= a;}
  mint operator*(const mint& a)const{ return mint(*this) *= a;}
  bool operator<(const mint& a)const{ return x < a.x;}
  bool operator==(const mint& a)const{ return x == a.x;}
};
istream& operator>>(istream&i,mint&a){i>>a.x;return i;}
ostream& operator<<(ostream&o,const mint&a){o<<a.x;return o;}
//



int main() {
  int n, m;
  cin >> n >> m;
  vector<string> a(n), b(m);
  rep(i, n) cin >> a[i];
  rep(i, m) cin >> b[i];
  rep(i, n-m) rep(j, n-m) {
    bool ok = true;
    rep(k, m) rep(l, m) {
      if (b[k][l] != a[i+k][j+l]) ok = false;
    }
    if (ok) {
      cout << "Yes\n";
      return 0;
    }
  }
  cout << "No\n";
}