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 city[4][100];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  if (k > (n - 2) * 2) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << endl;
  if (k % 2) {
    while (k > 0) {
      city[1][n / 2] = 1;
      --k;
      for (int j = 1; k > 0 && j < n / 2; ++j) {
        city[1][j + n / 2] = 1;
        city[1][n / 2 - j] = 1;
        k -= 2;
      }
      for (int j = 1; k > 0 && j <= n - 2; --k, ++j)
        if (j % 2)
          city[2][j / 2 + 1] = 1;
        else
          city[2][n - 1 - j / 2] = 1;
    }
  } else {
    for (int i = 1; i < 3; ++i)
      for (int j = 1; j <= k / 2; ++j) city[i][j] = 1;
  }
  for (int i = 0; i < 4; ++i) {
    for (int j = 0; j < n; ++j) {
      if (city[i][j] == 0)
        cout << '.';
      else
        cout << '#';
    }
    cout << endl;
  }
  return 0;
}
