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() {
  int n, k;
  cin >> n >> k;
  int m = k / 2;
  int r = k - m * 2;
  if (m > n - 3) {
    --m;
    r += 2;
  }
  char s[200];
  cout << "YES\n";
  for (int i = 0; i < 4; ++i) {
    for (int j = 0; j < n; ++j) {
      if (i == 0 || i == 3 || j == 0 || j == (n - 1)) {
        s[j] = '.';
      } else if (j < n / 2) {
        if (m > 0) {
          s[j] = '#';
          --m;
        } else {
          s[j] = '.';
        }
      } else if (j > n / 2) {
        s[j] = s[n - 1 - j];
      } else {
        if (r > 0) {
          s[j] = '#';
          --r;
        } else {
          s[j] = '.';
        }
      }
    }
    s[n] = '\0';
    cout << s << '\n';
  }
  return 0;
}
