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 = 1e5 + 100;
int n, k, dp[4][100];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n >> k;
  if (n % 2 == 0 && k % 2 == 0) return cout << "NO\n", 0;
  cout << "YES\n";
  if (k % 2 == 0) {
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) {
        if (j == 0 || j == n - 1)
          cout << '.';
        else if ((i == 1 || i == 2) && j <= k / 2)
          cout << '#';
        else
          cout << '.';
      }
      cout << '\n';
    }
    return 0;
  }
  vector<string> grid;
  for (int i = 0; i < 4; i++) grid.push_back(string(n, '.'));
  grid[1][n / 2] = '#';
  k--;
  int it = 1;
  while (k > 0 && it != n / 2) {
    grid[1][it] = grid[1][n - it - 1] = '#';
    k -= 2;
    ++it;
  }
  it = 1;
  while (k > 0 && it != n / 2) {
    grid[2][it] = grid[2][n - it - 1] = '#';
    k -= 2;
    ++it;
  }
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) cout << grid[i][j];
    cout << "\n";
  }
  return 0;
}
