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 mp[10][105];
int allc, targ;
int main() {
  scanf("%d%d", &allc, &targ);
  printf("YES\n");
  if (targ == 2 * (allc - 2) && allc != 2) mp[3][allc / 2 + 1] = '#';
  for (int j = 2; j < allc; ++j) {
    for (int i = 2; i <= 3; ++i) {
      if (targ <= 1) break;
      mp[i][j] = '#';
      mp[i][allc - j + 1] = '#';
      targ -= 2;
      if (targ <= 1) break;
    }
    if (targ <= 1) break;
  }
  if (targ == 1) mp[2][allc / 2 + 1] = '#';
  for (int i = 1; i <= 4; ++i) {
    for (int j = 1; j <= allc; ++j) {
      if (mp[i][j] == '#')
        printf("%c", mp[i][j]);
      else
        printf(".");
    }
    printf("\n");
  }
  return 0;
}
