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 + 10;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1);
const double eps = 1e-9;
using namespace std;
const long long mod = 1e9 + 7;
int ans, t, i, j, k, n, m;
int a[N];
char s[N][5];
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    for (int i = 1; i <= 4; i++) {
      for (int j = 1; j <= n; j++) {
        s[j][i] = '.';
      }
    }
    if (m % 2) {
      s[n / 2 + 1][3] = '#';
      m--;
    }
    {
      for (int i = 2; i < 4; i++) {
        if (m == 0) break;
        for (int j = 2; j < n;) {
          s[j][i] = '#';
          m--;
          if (j <= n / 2)
            j = (n - j) + 1;
          else
            j = (n - j) + 2;
          if (j == (n / 2 + 1) || m == 0) break;
        }
      }
      if (m) {
        for (int i = 2; i < 4; i++) {
          s[n / 2 + 1][i] = '#';
          m--;
          if (m == 0) break;
        }
      }
      printf("YES\n");
      for (int i = 1; i <= 4; i++) {
        for (int j = 1; j <= n; j++) {
          printf("%c", s[j][i]);
        }
        printf("\n");
      }
      continue;
    }
  }
  return 0;
}
