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;
const long long maxn = 1e5 + 500;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 500;
char jad[4][maxn];
int main() {
  long long n, k;
  cin >> n >> k;
  cout << "YES\n";
  for (long long i = 0; i < 4; i++) {
    for (long long j = 0; j < n; j++) {
      jad[i][j] = '.';
    }
  }
  if (k % 2 == 0) {
    for (long long j = 1; j <= k / 2; j++) {
      jad[1][j] = '#';
      jad[2][j] = '#';
    }
  }
  if (k == 1) {
    jad[1][n / 2] = '#';
  }
  if (k == 3) {
    jad[1][n / 2 - 1] = '#';
    jad[2][n / 2] = '#';
    jad[1][n / 2 + 1] = '#';
  }
  if (k % 2 == 1 && k > 4) {
    jad[1][2] = '#';
    jad[1][1] = '#';
    jad[2][1] = '#';
    for (long long j = 3; j <= k / 2 + 1; j++) {
      jad[1][j] = '#';
      jad[2][j] = '#';
    }
  }
  for (long long i = 0; i < 4; i++) {
    for (long long j = 0; j < n; j++) {
      cout << jad[i][j];
    }
    cout << endl;
  }
}
