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