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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n = 0, m = 0, k = 0, i = 0, j = 0, p = 0, q = 0, x = 0, y = 0,
                z = 0, ans = 0, cnt = 0;
  string s;
  bool flag = false;
  cin >> n >> k;
  cout << "YES" << endl;
  char op[2][n - 2];
  for (i = 0; i < n - 2; i++) op[0][i] = '.', op[1][i] = '.';
  if (k % 2 == 0) {
    for (j = 0; j < (k / 2); j++) op[0][j] = '#', op[1][j] = '#';
  } else {
    if (k > n - 2) {
      for (i = 0; i < n - 2; i++) op[0][i] = '#';
      k = k - (n - 2);
      j = 0;
      for (i = 0; i < k; i += 2)
        op[1][j] = '#', op[1][n - 2 - j - 1] = '#', j++;
    } else {
      op[0][(n - 2) / 2] = '#';
      j = 0;
      k--;
      for (i = 0; i < k; i += 2)
        op[0][j] = '#', op[0][n - 2 - j - 1] = '#', j++;
    }
  }
  for (i = 0; i < 4; i++) {
    for (j = 0; j < n; j++) {
      if (i == 0 || j == 0 || i == 3 || j == n - 1)
        cout << '.';
      else
        cout << op[i - 1][j - 1];
    }
    cout << endl;
  }
  return 0;
}
