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() {
  int n, k;
  cin >> n >> k;
  char a[4][n];
  for (int i = 0; i < 4; ++i) {
    for (int j = 0; j < n; ++j) {
      a[i][j] = '.';
    }
  }
  if (k <= 2 * (n - 2)) {
    if (!(k % 2)) {
      int j = 1;
      while (k) {
        a[1][j] = '#';
        a[2][j] = '#';
        j++;
        k -= 2;
      }
    } else {
      int i = 1;
      while (k > 0) {
        for (int l = 1; l < n - 1; l++) {
          a[i][l] = '#';
        }
        k = k - n + 2;
        i++;
      }
      i--;
      if (k != 0) {
        int j = 0;
        int mid = n / 2;
        if (i == 2) {
          k++;
          a[i][mid] = '.';
          while (k) {
            a[i][1 + j] = '.';
            a[i][n - 2 - j] = '.';
            k += 2;
            j++;
          }
        } else {
          while (k) {
            a[i][1 + j] = '.';
            a[i][n - 2 - j] = '.';
            k += 2;
            j++;
          }
        }
      }
    }
    cout << "YES" << endl;
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) {
        cout << a[i][j];
      }
      cout << endl;
    }
  } else {
    cout << "NO" << endl;
  }
}
