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() {
  long long n, k;
  cin >> n >> k;
  char a[5][n + 1];
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= n; j++) {
      a[i][j] = '.';
    }
  }
  if (n == 3 && k == 2) {
    cout << "YES\n";
    cout << "...\n.#.\n.#.\n...";
    return 0;
  }
  long long t = k;
  if (k % 2 == 1) {
    a[2][(n / 2) + 1] = '#';
    k--;
  }
  long long st = 2, en = n - 1;
  while (k > 0 && st <= en) {
    if (st != en)
      k -= 2;
    else {
      if (t % 2 == 0) {
        break;
      }
      if (a[2][st] != '#') {
        k--;
      }
    }
    a[2][st] = '#';
    a[2][en] = '#';
    st++;
    en--;
  }
  st = 2, en = n - 1;
  while (k > 0 && st <= en) {
    if (st != en)
      k -= 2;
    else {
      if (a[3][st] != '#') {
        k--;
      }
    }
    a[3][st] = '#';
    a[3][en] = '#';
    st++;
    en--;
  }
  if (k > 0) {
    if (a[3][(n / 2) + 1] != '#') k--;
    a[3][(n / 2) + 1] = '#';
  }
  if (k > 0) {
    a[2][(n / 2) + 1] = '#';
    k--;
  }
  cout << "YES\n";
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= n; j++) {
      cout << a[i][j];
    }
    cout << "\n";
  }
  return 0;
}
