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>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string s[2];
  long long int n, k, i;
  cin >> n >> k;
  for (i = 0; i < n; i++) {
    s[0] += ".";
    s[1] += ".";
  }
  if (k % 2 == 0) {
    for (i = 1; i <= k / 2; i++) s[0][i] = s[1][i] = '#';
  } else {
    if (k <= n - 2) {
      k--;
      s[0][n / 2] = '#';
      for (i = 1; i <= k / 2; i++) s[0][n / 2 - i] = s[0][n / 2 + i] = '#';
    } else {
      for (i = 1; i <= n - 2; i++) s[0][i] = '#';
      k -= n - 2;
      for (i = 1; i <= k / 2; i++) s[1][i] = s[1][n - 1 - i] = '#';
    }
  }
  cout << "YES\n";
  for (i = 0; i < n; i++) cout << ".";
  cout << "\n";
  cout << s[0] << "\n";
  cout << s[1] << "\n";
  for (i = 0; i < n; i++) cout << ".";
  cout << "\n";
  return 0;
}
