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;
  vector<vector<int>> x(4, vector<int>(n));
  if (k > (n - 2) * 2)
    cout << "NO";
  else {
    cout << "YES" << endl;
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) {
        if (i == 0 || j == 0 || i == 3 || j == n - 1)
          x[i][j] = 0;
        else {
          if (j != n / 2 && k > 1 && x[i][j] == 0) {
            x[i][j] = 1;
            x[i][n - j - 1] = 1;
            k -= 2;
          }
        }
      }
    }
    int i = 1;
    while (k > 0) {
      x[i][n / 2] = 1;
      --k;
      i++;
    }
    for (i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) {
        if (x[i][j] == 0)
          cout << '.';
        else
          cout << '#';
      }
      cout << endl;
    }
    cin >> n;
  }
  return (0);
}
