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>
const int N = 1e2 + 9;
using namespace std;
int n, m;
int mp[N][N];
int main() {
  scanf("%d%d", &n, &m);
  puts("YES");
  memset(mp, 0, sizeof mp);
  if (m <= n - 2) {
    int mid = (n + 1) / 2;
    if (m & 1) {
      mp[2][mid] = 1;
    }
    for (int i = 1; i <= m / 2; i++) {
      mp[2][mid - i] = mp[2][mid + i] = 1;
    }
  } else {
    for (int i = 2; i < n; i++) {
      mp[2][i] = 1;
    }
    m -= n - 2;
    int mid = (n + 1) / 2;
    if (m & 1) {
      mp[3][mid] = 1;
    }
    for (int i = 1; i <= m / 2; i++) {
      mp[3][mid - i] = mp[3][mid + i] = 1;
    }
  }
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= n; j++) {
      if (mp[i][j] == 1)
        printf("#");
      else
        printf(".");
    }
    puts("");
  }
}
