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;
bool vec[8][105];
int n, k;
char c[] = ".#";
void Afisare() {
  int i, j;
  for (i = 1; i <= 4; i++) {
    for (j = 1; j <= n; j++) {
      cout << c[vec[i][j]];
    }
    cout << "\n";
  }
}
int main() {
  int i, cont = 0;
  cin >> n >> k;
  if (k % 2 == 0) {
    for (i = 2; i <= k / 2 + 1; i++) {
      vec[2][i] = 1;
      vec[3][i] = 1;
    }
  } else if (k <= n - 2) {
    for (i = (n - k) / 2 + 1; i < (n - k) / 2 + 1 + k; i++) {
      if (i % 2 == 0)
        vec[2][i] = 1;
      else
        vec[3][i] = 1;
    }
  } else {
    for (i = 2; i <= n - 1; i++) {
      vec[2][i] = 1;
      k--;
    }
    for (i = 2; i <= n - 1 && k > 1; i++) {
      vec[3][i] = 1;
      k--;
      cont++;
    }
    vec[3][n - 1] = 1;
  }
  cout << "YES\n";
  Afisare();
  return 0;
}
