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;
using ll = long long;
vector<string> s(100, string(100, '.'));
int n, k;
int solve() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n >> k;
  if (n - 2 == k) {
    for (int i = 1; i < n - 1; i++) {
      s[1][i] = '#';
    }
    return 1;
  }
  if (k % 2 == 0) {
    for (int j = 0; j < n; j++) {
      for (int i = 0; i < 4; i++) {
        if (i != 0 && i != 3 && j != 0 && j != n - 1 && k) {
          s[i][j] = '#';
          k--;
        }
      }
    }
    return 1;
  }
  if (k >= 5) {
    s[1][1] = '#';
    s[2][1] = '#';
    k -= 2;
    for (int i = 2; i < n - 2; i++) {
      s[1][i] = '#';
      k--;
      if (k == 2) {
        s[1][i + 1] = s[2][i + 1] = '#';
        return 1;
      }
    }
    s[1][n - 2] = s[2][n - 2] = '#';
    k -= 2;
    for (int i = 2; i < n - 2; i++) {
      s[2][i] = '#';
      k--;
      if (k == 0) return 1;
    }
  }
  if (k == 3) {
    s[1][n / 2 + 1] = s[2][n / 2] = s[1][n / 2 - 1] = '#';
    return 1;
  }
  if (k == 1) {
    s[2][n / 2] = '#';
    return 1;
  }
  assert(0);
}
int main() {
  int t = solve();
  if (!solve()) return (cout << "NO" << endl, 0);
  cout << "YES" << endl;
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) {
      cout << s[i][j];
    }
    cout << endl;
  }
}
