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 long long INF = (int)1e9 + 7;
const int mod = (int)1e9 + 7;
const int N = (int)2e5 + 2;
int n, k;
char res[105][105];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= 4; ++i) {
    for (int j = 1; j <= n; ++j) res[i][j] = '.';
  }
  cout << "YES\n";
  if (k == (n - 2) * 2) {
    for (int i = 2; i <= n - 1; ++i) res[2][i] = res[3][i] = '#';
    for (int i = 1; i <= 4; ++i) {
      for (int j = 1; j <= n; ++j) cout << res[i][j];
      cout << "\n";
    }
    return 0;
  }
  if (k & 1) res[2][(n + 1) / 2] = '#', k--;
  int x = 2, l = 2, r = n - 1;
  while (k) {
    if (l == r) {
      x++;
      l = 2, r = n - 1;
    }
    res[x][l] = '#';
    res[x][r] = '#';
    l++, r--;
    k -= 2;
  }
  for (int i = 1; i <= 4; ++i) {
    for (int j = 1; j <= n; ++j) cout << res[i][j];
    cout << "\n";
  }
  return 0;
}
