The city of Fishtopia can be imagined as a grid of 4 rows and an odd number of columns. It has two main villages; the first is located at the top-left cell (1,1), people who stay there love fishing at the Tuna pond at the bottom-right cell (4, n). The second village is located at (4, 1) and its people love the Salmon pond at (1, n).

The mayor of Fishtopia wants to place k hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells.

A person can move from one cell to another if those cells are not occupied by hotels and share a side.

Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond?

Input

The first line of input contain two integers, n and k (3 ≤ n ≤ 99, 0 ≤ k ≤ 2×(n-2)), n is odd, the width of the city, and the number of hotels to be placed, respectively.

Output

Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO".

If it is possible, print an extra 4 lines that describe the city, each line should have n characters, each of which is "#" if that cell has a hotel on it, or "." if not.

Examples

Input

7 2


Output

YES
.......
.#.....
.#.....
.......


Input

5 3


Output

YES
.....
.###.
.....
.....
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
stringstream output;
inline void eOP();
inline void solve() {
  output << "YES\n";
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < 4; ++i) {
    for (int j = 0; j < n; ++j) {
      if (i == 0 || i == 3 || j == 0 || j == n - 1) {
        output << ".";
        continue;
      }
      if (j == (n >> 1)) {
        if (i == 1) {
          if (k) {
            output << "#";
            continue;
          } else {
            output << ".";
            continue;
          }
        } else {
          if (k % 2 == 0 && k) {
            output << "#";
            continue;
          } else {
            output << ".";
            continue;
          }
        }
      }
      int x = min(j, n - 1 - j);
      if (x + (n / 2 - 1) * (i == 2) <= (k - 1) / 2) {
        output << "#";
        continue;
      } else {
        output << ".";
        continue;
      }
    }
    output << "\n";
  }
  output << flush;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  size_t t = 1;
  for (int i = 0; i < t; ++i) {
    solve();
  }
  eOP();
}
inline void eOP() {
  cout << output.str();
  output.str(string());
  exit(0);
}
