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
// ------------------------------------------------
#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
#include <math.h>

using namespace std;

// func
// ------------------------------------------------
int CalcSumOfDigit(int n);      // 各桁の和を計算する。
string upper(string str);       // 英字を大文字に変換する。
string lower(string str);       // 英字を小文字に変換する。

// define
// ------------------------------------------------
#define all(a)        (a).begin(),(a).end()
#define rall(a)       (a).rbegin(),(a).rend()
#define pb            push_back
#define sz(a)         int((a).size())
#define rep(i,n)      for(int(i)=0;(i)<(n);(i)++)
#define repe(i,n)     for(int(i)=0;(i)<=(n);(i)++)
#define vsort(v)      sort((v).begin(),(v).end())
#define rvsort(v)     sort(rall((v)))
#define vi            vector<int>
#define GCD(a,b)      __gcd((a),(b))
#define LCM(a,b)      (a)/GCD((a),(b))*(b)
#define kiriage(a,b)  ((a)+(b)-1)/(b)

const int INF = 1e9;

// code
// ------------------------------------------------
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];

  bool ans = false;
  rep(i,n)
  {
    int pos = a[i].find(b[0]);
    if(pos == -1)
    {
      continue;
    }
    bool _is = true;
    rep(j,m)
    {
      if(i + j >= n)
      {
        break;
      }
      string buf = a[i + j].substr(pos, sz(b[j]));
      if(buf != b[j])
      {
        _is = false;
      }
    }
    if(_is)
    {
      ans = true;
      break;
    }
  }

  if(ans) cout << "Yes" << endl;
  else cout << "No" << endl;

  return 0;
}
// funcの実体
// ------------------------------------------------
int CalcSumOfDigit(int n)
{
  int s = 0;
  while(n)
  {
    s += n % 10;
    n = n / 10;
  }
  return s;
}

string upper(string str)
{
  for(auto itr = str.begin();itr != str.end() ; itr++)
  {
    if(97 <= *itr && *itr <= 122)
    {
      *itr = *itr - 32;
    }
  }
  return str;
}

string lower(string str)
{
  for(auto itr = str.begin();itr != str.end() ; itr++)
  {
    if(65 <= *itr && *itr <= 90)
    {
      *itr = *itr + 32;
    }
  }
  return str;
}