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() {
  int n, k;
  std::cin >> n >> k;
  if (k == 0) {
    std::cout << "YES" << '\n';
    std::cout << std::string(n, '.') << '\n';
    std::cout << std::string(n, '.') << '\n';
    std::cout << std::string(n, '.') << '\n';
    std::cout << std::string(n, '.') << '\n';
    return 0;
  }
  if (k == n - 2) {
    std::cout << "YES" << '\n';
    std::cout << std::string(n, '.') << '\n';
    std::cout << '.' << std::string(n - 2, '#') << '.' << '\n';
    std::cout << std::string(n, '.') << '\n';
    std::cout << std::string(n, '.') << '\n';
    return 0;
  }
  if (k % 2 == 0) {
    std::cout << "YES" << '\n';
    std::cout << std::string(n, '.') << '\n';
    std::cout << '.' << std::string(k / 2, '#')
              << std::string(n - 1 - k / 2, '.') << '\n';
    std::cout << '.' << std::string(k / 2, '#')
              << std::string(n - 1 - k / 2, '.') << '\n';
    std::cout << std::string(n, '.') << '\n';
    return 0;
  } else if (k < n - 2) {
    std::cout << "YES" << '\n';
    std::cout << std::string(n, '.') << '\n';
    std::cout << std::string((n - k) / 2, '.') << std::string(k, '#')
              << std::string((n - k) / 2, '.') << '\n';
    std::cout << std::string(n, '.') << '\n';
    std::cout << std::string(n, '.') << '\n';
    return 0;
  } else {
    std::cout << "YES" << '\n';
    std::cout << std::string(n, '.') << '\n';
    std::cout << '.' << std::string(n - 2, '#') << '.' << '\n';
    k -= n - 1;
    std::cout << '.' << std::string(k, '#') << std::string(n - 3 - k, '.')
              << '#' << '.' << '\n';
    std::cout << std::string(n, '.') << '\n';
    return 0;
  }
  return 0;
}
