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 main() {
  int n, k;
  char s[100][100];
  scanf("%d%d", &n, &k);
  for (int i = 0; i < 4; i++)
    for (int j = 0; j < n; j++) s[i][j] = '.';
  if (k == 0) {
    printf("YES\n");
  } else {
    if (k % 2 == 0) {
      for (int i = 1; i < n - 1 && k > 0; i++) {
        s[1][i] = '#';
        s[2][i] = '#';
        k -= 2;
      }
      printf("YES\n");
    } else {
      s[1][n / 2] = '#';
      k--;
      for (int i = 1; i < (n / 2) && k > 0; i++) {
        s[1][(n / 2) + i] = '#';
        s[1][(n / 2) - i] = '#';
        k -= 2;
      }
      for (int i = 1; i < (n / 2) && k > 0; i++) {
        s[2][(n / 2) + i] = '#';
        s[2][(n / 2) - i] = '#';
        k -= 2;
      }
      printf("YES\n");
    }
  }
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) printf("%c", s[i][j]);
    printf("\n");
  }
  return 0;
}
