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;
char c[9][109];
int n, k, a, m;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= n; j++) c[i][j] = '.';
  if (k == 3) {
    c[2][2] = '#';
    c[2][(n + 1) / 2] = '#';
    c[2][n - 1] = '#';
  } else if (k == 1) {
    c[2][(n + 1) / 2] = '#';
  } else {
    m = (k + 1) / 2;
    for (int i = 2; i < 2 + m; i++) {
      c[2][i] = '#';
      c[3][i] = '#';
    }
    if (k % 2) c[2][m] = '.';
  }
  cout << "YES\n";
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= n; j++) cout << c[i][j];
    cout << endl;
  }
}
