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>
char ans[5][105];
int main() {
  register int n, k;
  scanf("%d%d", &n, &k);
  printf("YES\n");
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 105; j++) ans[i][j] = '.';
  if (k % 2 == 0) {
    int p = 2;
    while (k > 0) {
      ans[2][p] = '#';
      ans[3][p] = '#';
      p++;
      k -= 2;
    }
  } else {
    if (k == 1)
      ans[2][n / 2 + 1] = '#';
    else if (k == 3) {
      int s = (n + 1) / 2;
      ans[2][s] = '#';
      ans[2][s - 1] = '#';
      ans[2][s + 1] = '#';
    } else {
      ans[2][2] = '#';
      ans[2][3] = '#';
      ans[3][2] = '#';
      register int p = 4;
      k -= 3;
      while (k > 0) {
        k -= 2;
        ans[2][p] = '#';
        ans[3][p] = '#';
        p++;
      }
    }
  }
  for (register int i = 1; i < 5; i++, puts(""))
    for (register int j = 1; j <= n; j++) putchar(ans[i][j]);
  return 0;
}
