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>
int main() {
  unsigned int n, k;
  std::cin >> n >> k;
  std::vector<std::vector<char>> mat(4, std::vector<char>(n, '.'));
  if (k <= 2 * (n - 2)) {
    std::cout << "YES" << std::endl;
    if (k % 2 == 0) {
      for (int j = 1; j < n - 1 && k > 0; j++) {
        mat[1][j] = '#';
        mat[2][j] = '#';
        k -= 2;
      }
    } else {
      int first = k;
      int second = 0;
      if (k > n - 2) {
        first = n - 2;
        second = k - first;
      }
      if (first) {
        for (int j = n / 2 - first / 2; k > 0 && (j <= n / 2 + first / 2);
             j++) {
          mat[1][j] = '#';
          k--;
        }
      }
      if (second) {
        if (second % 2 == 0) {
          mat[2][n / 2 + (n - 2) / 2] = '#';
          k--;
        }
        for (int j = 1; k > 0 && j < n - 2; j++) {
          mat[2][j] = '#';
          k--;
        }
      }
    }
    for (auto i = 0; i < 4; i++) {
      for (auto j = 0; j < n; j++) {
        std::cout << mat[i][j];
      }
      std::cout << std::endl;
    }
  } else {
    std::cout << "NO";
  }
  return 0;
}
