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;
using ll = long long;
using ull = unsigned long long;
const int MAXN = 999 + 9, INF = 0x3f3f3f3f, mod = 1e9 + 7;
const ll INF_ll = 1e18, mod_l = 1e9 + 7;
const long double PI = 3.14159265358979, eps = 1e-9;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  puts("YES");
  char ans[MAXN][2];
  memset(ans, '.', sizeof(ans));
  if (k & 1) {
    int mid = n / 2 + 1;
    ans[mid][0] = '#';
    k--;
    for (int i = 1; i < n / 2 && k; i++) {
      ans[mid + i][0] = '#';
      ans[mid - i][0] = '#';
      k -= 2;
    }
    for (int i = 1; i < n / 2 && k; i++) {
      ans[mid + i][1] = '#';
      ans[mid - i][1] = '#';
      k -= 2;
    }
  } else {
    for (int j = 2; k; k -= 2, j++) {
      for (int i = 0; i < 2; i++) {
        ans[j][i] = '#';
      }
    }
  }
  for (int i = 0; i < n; i++) putchar('.');
  putchar(10);
  for (int j = 0; j < 2; j++) {
    for (int i = 1; i <= n; i++) putchar(ans[i][j]);
    putchar(10);
  }
  for (int i = 0; i < n; i++) putchar('.');
  putchar(10);
  return 0;
}
