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 n, k;
char a[1000][1000];
void print() {
  puts("YES");
  fflush(stdout);
  for (int i = (1); i <= (4); ++i) {
    for (int j = (1); j <= (n); ++j) {
      if (a[i][j] != '#')
        cout << ".";
      else
        cout << "#";
    }
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin.precision(10);
  cout << fixed;
  cin >> n >> k;
  int tmp = k;
  int i = 2, j1 = 2, j2 = n - 1, m = n / 2 + 1;
  if (k % 2) a[2][m] = '#', --k;
  while (k > 0) {
    a[i][j1] = a[i][j2] = '#';
    if (j1 == j2)
      --k;
    else
      k -= 2;
    ++j1;
    --j2;
    if ((j1 == m && j2 == m) || (j1 > j2)) {
      j1 = 2;
      j2 = n - 1;
      ++i;
    }
    if (i == 4) break;
  }
  if (tmp == (2 * (n - 2))) a[2][m] = a[3][m] = '#';
  print();
  return 0;
}
