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 a[4][100];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) a[i][j] = '.';
  }
  if (k > 2 * (n - 2))
    printf("NO\n");
  else {
    printf("YES\n");
    if (k % 2 == 0) {
      for (int i = 1; i < n - 1; i++) {
        for (int j = 1; j < 3; j++) {
          if (k > 0) {
            k--;
            a[j][i] = '#';
          }
        }
      }
    } else {
      a[1][n / 2] = '#';
      k--;
      if (k > 0) {
        for (int j = 1; j < n / 2; j++) {
          if (k > 0) {
            a[1][j] = '#';
            a[1][n - j - 1] = '#';
            k = k - 2;
          }
        }
      }
      if (k > 0) {
        for (int j = 1; j < n / 2; j++) {
          if (k > 0) {
            a[2][j] = '#';
            a[2][n - j - 1] = '#';
            k = k - 2;
          }
        }
      }
    }
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) printf("%c", a[i][j]);
      printf("\n");
    }
  }
}
