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() {
  char s[4][100];
  for (int i = 0; i < 4; i++) {
    for (long long int j = 1; j < 100; j++) s[i][j] = '.';
  }
  long long int n, k;
  cin >> n >> k;
  cout << "YES\n";
  if (k != 0) {
    if (k % 2 == 0) {
      long long int i = 1, j = n / 2 + 1;
      s[1][j] = '#';
      s[2][j] = '#';
      j--;
      k -= 2;
      while (k != 0) {
        s[i][j] = '#';
        s[i][n - j + 1] = '#';
        k -= 2;
        j--;
        if (j == 1) {
          i = 2;
          j = n / 2;
        }
      }
    } else {
      long long int j = n / 2 + 1;
      if (k <= n - 2) {
        s[1][j] = '#';
        j--;
        k--;
        while (k != 0) {
          s[1][j] = '#';
          s[1][n - j + 1] = '#';
          j--;
          k -= 2;
        }
      } else {
        for (long long int i = 2; i < n; i++) s[1][i] = '#';
        k -= n - 2;
        if (k % 2 == 0) {
          long long int j = n / 2;
          while (k != 0) {
            s[2][j] = '#';
            s[2][n - j + 1] = '#';
            j--;
            k -= 2;
          }
        } else {
          long long int j = n / 2 + 1;
          s[2][j] = '#';
          k--;
          j--;
          while (k != 0) {
            s[2][j] = '#';
            s[2][n - j + 1] = '#';
            j--;
            k -= 2;
          }
        }
      }
    }
  }
  for (long long int i = 0; i < 4; i++) {
    for (long long int j = 1; j <= n; j++) cout << s[i][j];
    cout << "\n";
  }
  return 0;
}
