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;
string a[4];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  puts("YES");
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) a[i].push_back('.');
  }
  if (k % 2 == 0) {
    int c = 1;
    int l = 1;
    while (1) {
      if (k <= 0) break;
      a[c][l] = '#';
      c++;
      k--;
      if (k <= 0) break;
      if (c >= 3) {
        c = 1;
        l++;
      }
    }
  } else {
    int l = n / 2, r = n / 2;
    int c = 1;
    while (1) {
      if (k <= 0) break;
      a[c][l] = a[c][r] = '#';
      if (l == r)
        k--;
      else
        k -= 2;
      if (k <= 0) break;
      l--, r++;
      if (l <= 0) break;
    }
    l = n / 2 - 1, r = n / 2 + 1;
    c = 2;
    while (1) {
      if (k <= 0) break;
      a[c][l] = a[c][r] = '#';
      k -= 2;
      l--, r++;
    }
  }
  for (int i = 0; i < 4; i++) cout << a[i] << endl;
  return 0;
}
