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;
char a[5][(100 + 10)] = {0};
int main() {
  int n, y;
  cin >> n >> y;
  for (int i = 1; i <= 4; i++)
    for (int j = 1; j <= n; j++) a[i][j] = '.';
  if (y % 2 == 0) {
    int p = y / 2;
    for (int i = (n - p) / 2 + 1; i <= (n - p) / 2 + p; i++)
      a[2][i] = a[3][i] = '#';
  } else {
    if (y <= n - 2) {
      int p = y;
      for (int i = (n - p) / 2 + 1; i <= (n - p) / 2 + p; i++) a[2][i] = '#';
    } else {
      for (int i = 2; i <= n - 1; i++) a[2][i] = '#';
      int p = (y - n + 2) / 2;
      for (int i = 1; i <= p; i++) a[3][1 + i] = a[3][n - i] = '#';
    }
  }
  cout << "YES" << endl;
  for (int i = 1; i <= 4; i++) cout << a[i] + 1 << endl;
  return 0;
}
