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;
const int maxn = 1e2 + 10;
const int mod = 1e9 + 7;
int n, k, m, w;
char a[4][1000];
int main() {
  cin >> n >> k;
  m = n / 2;
  w = k / 2;
  k %= 2;
  cout << "YES\n";
  for (int q = 1; q < 3; ++q) {
    for (int j = 1, i = n - 2; j < n / 2; ++j, --i) {
      if (w > 0) {
        w--;
        a[q][i] = a[q][j] = '#';
      }
    }
  }
  k += (w * 2);
  for (int i = 1; i < 3; ++i) {
    if (k > 0) {
      k--;
      a[i][m] = '#';
    }
  }
  for (int i = 0; i < 4; ++i) {
    for (int j = 0; j < n; ++j) {
      if (a[i][j] == '#')
        cout << a[i][j];
      else
        cout << '.';
    }
    cout << '\n';
  }
  return 0;
}
