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 main() {
  long long int n, k;
  cin >> n >> k;
  if (k % 2 == 0) {
    cout << "YES\n";
    for (long long int i = 0; i < 4; i++) {
      for (long long int j = 0; j < n; j++) {
        if ((i == 1 || i == 2) && (j > 0 && (j <= (k / 2)))) {
          cout << "#";
        } else {
          cout << ".";
        }
      }
      cout << "\n";
    }
  } else {
    if (k > n - 2) {
      cout << "YES\n";
      for (long long int i = 0; i < 4; i++) {
        for (long long int j = 0; j < n; j++) {
          if (i == 1 && (j > 0 && j < n - 1)) {
            cout << "#";
            continue;
          }
          if ((i == 2) &&
              ((j > 0 && j < n - 1) &&
               (j <= (k - n + 2) / 2 || j >= (n + n + n - k - 2) / 2 - 1))) {
            cout << "#";
          } else {
            cout << ".";
          }
        }
        cout << "\n";
      }
    } else {
      cout << "YES\n";
      for (long long int i = 0; i < 4; i++) {
        for (long long int j = 0; j < n; j++) {
          if ((i == 1) && ((j >= ((n - k) / 2)) && (j < ((n + k) / 2)))) {
            cout << "#";
          } else {
            cout << ".";
          }
        }
        cout << "\n";
      }
    }
  }
  return 0;
}
