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 long long INFLL = 2 * (long long)1e18 + 100;
const int INFINT = 2 * (int)1e9 + 100;
const double PI = atan(1) * 4;
const double EPS = 1e-9;
const int SEED = 1e3 + 7;
const int MOD = 1e9 + 9;
const int NMAX = 1e6 + 5;
char mat[200][200];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  if (m % 2 == 0)
    for (int j = 2; j <= n - 1; j++)
      for (int i = 2; i <= 3; i++) {
        if (m)
          mat[i][j] = '#';
        else
          break;
        m--;
      }
  else {
    if (m % 2 && m < n - 2) {
      mat[2][n / 2 + 1] = '#';
      m--;
      int cnt = 0;
      for (int j = n / 2; j >= 2; j--) {
        if (cnt < m / 2)
          cnt++;
        else
          break;
        mat[2][j] = '#';
      }
      cnt = 0;
      for (int j = n / 2 + 2; j <= n - 1; j++) {
        if (cnt < m / 2)
          cnt++;
        else
          break;
        mat[2][j] = '#';
      }
      goto SKIP;
    }
    for (int j = 2; j <= n - 1; j++) {
      mat[2][j] = '#';
      m--;
    }
    if (m) {
      mat[3][2] = '#';
      mat[3][n - 1] = '#';
      m -= 2;
      for (int j = 3; j < n - 1; j++) {
        if (m)
          mat[3][j] = '#';
        else
          break;
        m--;
      }
    }
  }
SKIP:;
  {
    cout << "YES\n";
    for (int i = 1; i <= 4; i++)
      for (int j = 1; j <= n; j++) {
        if (mat[i][j] != '#')
          cout << ".";
        else
          cout << mat[i][j];
        if (j == n) cout << '\n';
      }
  }
  return 0;
}
