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)

#include <bits/stdc++.h>
#define fi first
#define se second
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rrep(i, n) for (int i = 1; i <= (n); ++i)
#define drep(i, n) for (int i = (n)-1; i >= 0; --i)
#define srep(i, s, t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(), a.end()
#define rrng(a) a.rbegin(), a.rend()
#define maxs(x, y) (x = max(x, y))
#define mins(x, y) (x = min(x, y))
#define limit(x, l, r) max(l, min(x, r))
#define lims(x, l, r) (x = max(l, min(x, r)))
#define isin(x, l, r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define uni(x) x.erase(unique(rng(x)), x.end())
#define show(x) cout << #x << " = " << x << endl;
#define PQ(T) priority_queue<T, v(T), greater<T>>
#define bn(x) ((1 << x) - 1)
#define dup(x, y) (((x) + (y)-1) / (y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef tuple<int, int, int> T;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
typedef vector<T> vt;

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];
  }
  int isMatch = 1;
  if(n==m){
    rep(i,n)if(a[i] != b[i]) isMatch = 0;
    if(isMatch){
      cout<<"Yes"<<endl;
      return 0;
    }
  }

  rep(i,n){
    rep(j,n){
      isMatch = 1;
      if(i > n-m || j > n-m) continue;
      rep(k,m){
        string s = a[i+k].substr(j,m);
        if(s != b[k])isMatch=0;
      }
      if(isMatch){
        cout<<"Yes"<<endl;
        return 0;
      }
    }
  }
  cout<<"No"<<endl;
  return 0;
}