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;
int main() {
  int n, k;
  char ch[4][102];
  for (int i = 0; i < 100; i++) ch[0][i] = ch[3][i] = '.';
  ch[1][0] = ch[2][0] = '.';
  while (cin >> n >> k) {
    bool flag = true;
    if (k <= n - 2 || k % 2 == 0) {
      if (k % 2 == 0) {
        k = k / 2;
        for (int i = 1; i < n; i++) {
          if (i <= k)
            ch[1][i] = ch[2][i] = '#';
          else
            ch[1][i] = ch[2][i] = '.';
        }
      } else {
        if ((n - k) % 2 == 0) {
          int x = (n - k) / 2;
          for (int i = 1; i < n; i++) {
            if (i < x || i >= (n - x))
              ch[1][i] = '.';
            else
              ch[1][i] = '#';
          }
          for (int i = 1; i < n; i++) ch[2][i] = '.';
        } else {
          flag = false;
        }
      }
    } else {
      if (k - n >= 0) {
        for (int i = 1; i < n; i++) {
          if (i == n - 1)
            ch[1][i] = '.';
          else
            ch[1][i] = '#';
        }
        k = k - n + 1;
        int x = n - 2;
        for (int i = 1; i < n; i++) {
          if (i == x || k-- > 0) {
            ch[2][i] = '#';
          } else
            ch[2][i] = '.';
        }
      } else
        flag = false;
    }
    if (flag) {
      printf("YES\n");
      for (int i = 0; i < 4; i++) {
        for (int j = 0; j < n; j++) printf("%c", ch[i][j]);
        printf("\n");
      }
    } else
      printf("NO\n");
  }
  return 0;
}
