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 int N = 1e5 + 5;
int T, n, m;
char grid[6][105];
int main() {
  scanf("%d %d", &n, &m);
  if (m % 2) {
    int md = n / 2;
    grid[1][md] = '#';
    --m;
    int dis = 1;
    while (m && md + dis < n - 1) {
      grid[1][md + dis] = '#';
      grid[1][md - dis] = '#';
      m -= 2;
      ++dis;
    }
    dis = 1;
    while (m && md + dis < n - 1) {
      grid[2][md + dis] = '#';
      grid[2][md - dis] = '#';
      m -= 2;
      ++dis;
    }
  } else {
    int x = 1;
    while (m) {
      grid[1][x] = '#';
      grid[2][x] = '#';
      m -= 2;
      ++x;
    }
  }
  printf("YES\n");
  for (int i = 0; i < 4; ++i) {
    for (int x = 0; x < n; ++x) {
      if (grid[i][x] == '#')
        printf("#");
      else
        printf(".");
    }
    printf("\n");
  }
}
