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;
char tab[5][110];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = int(0); i < int(4); ++i)
    for (int j = int(0); j < int(n); ++j) tab[i][j] = '.';
  if (k & 1) {
    int mid = n >> 1;
    tab[1][mid] = '#';
    k--;
    for (int t = int(1); t < int(3); ++t) {
      if (k == 0) break;
      int i = mid - 1, j = mid + 1;
      while (i >= 1 and k) {
        tab[t][i] = tab[t][j] = '#';
        k -= 2;
        --i;
        ++j;
      }
    }
    cout << "YES" << endl;
    for (int i = int(0); i < int(4); ++i) {
      for (int j = int(0); j < int(n); ++j) cout << tab[i][j];
      cout << endl;
    }
  } else {
    for (int i = 1; i < n and k; ++i) {
      tab[1][i] = tab[2][i] = '#';
      k -= 2;
    }
    cout << "YES" << endl;
    for (int i = int(0); i < int(4); ++i) {
      for (int j = int(0); j < int(n); ++j) cout << tab[i][j];
      cout << endl;
    }
  }
  return 0;
}
