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 N, K;
bool C[4][100];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> K;
  for (int i = 2; K; i++) {
    if (K >= 4) {
      C[1][i - 1] = 1;
      C[2][i - 1] = 1;
      C[1][N - i] = 1;
      C[2][N - i] = 1;
      K -= 4;
    } else if (K >= 2) {
      C[1][i - 1] = 1;
      C[1][N - i] = 1;
      if (i - 1 == N - i) {
        C[2][i - 1] = 1;
      }
      K -= 2;
    } else {
      C[1][N / 2] = 1;
      K--;
    }
  }
  cout << "YES\n";
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < N; j++) {
      if (C[i][j]) {
        cout << '#';
      } else {
        cout << '.';
      }
    }
    cout << '\n';
  }
  return 0;
}
