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 N = 2e5 + 10;
char mp[300][300];
int main() {
  std::ios::sync_with_stdio(false);
  int n, k;
  while (cin >> n >> k) {
    for (int i = 1; i <= 4; i++) {
      for (int j = 1; j <= n; j++) {
        mp[i][j] = '.';
      }
    }
    if ((k & 1) == 0) {
      cout << "YES" << endl;
      for (int i = 2; i <= n - 1; i++) {
        if (k) {
          mp[2][i] = mp[3][i] = '#';
          k -= 2;
        }
      }
    } else {
      if (k >= 5) {
        cout << "YES" << endl;
        for (int i = 2; i <= n - 1; i++) {
          if (i == 3) {
            mp[2][i] = '#';
            k--;
            continue;
          }
          if (k) {
            mp[2][i] = mp[3][i] = '#';
            k -= 2;
          }
        }
      } else if (k <= n - 2) {
        cout << "YES" << endl;
        mp[2][n / 2 + 1] = '#';
        k--;
        int i = 1;
        while (k) {
          mp[2][n / 2 + 1 + i] = mp[2][n / 2 + 1 - i] = '#';
          k -= 2;
          i++;
        }
      } else {
        cout << "NO" << endl;
        continue;
      }
    }
    for (int i = 1; i <= 4; i++) {
      for (int j = 1; j <= n; j++) {
        cout << mp[i][j];
      }
      cout << endl;
    }
  }
}
