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, t;
int adadsasd[101];
int main() {
  cin >> n >> k;
  if (k % 2 == 0) {
    cout << "YES\n";
    for (int i = 1; i <= n; i++) cout << '.';
    cout << endl;
    cout << '.';
    t = k / 2 + k % 2;
    k -= t;
    for (int i = 2; i < n; i++) {
      if (t > 0) {
        cout << '#';
        t--;
      } else
        cout << '.';
    }
    cout << '.' << endl << '.';
    for (int i = 2; i < n; i++)
      if (k > 0) {
        cout << '#';
        k--;
      } else
        cout << '.';
    cout << '.' << endl;
    for (int i = 1; i <= n; i++) cout << '.';
    return 0;
  }
  if (k > n - 2) {
    cout << "YES\n";
    for (int i = 1; i <= n; i++) cout << '.';
    cout << endl;
    cout << '.';
    for (int i = 1; i <= n - 2; i++) cout << '#';
    cout << '.' << endl;
    k -= (n - 2);
    k /= 2;
    cout << '.';
    for (int i = 1; i <= k; i++) cout << '#';
    for (int i = 1; i <= n - 2 - 2 * k; i++) cout << '.';
    for (int i = 1; i <= k; i++) cout << '#';
    cout << '.';
    cout << endl;
    for (int i = 1; i <= n; i++) cout << '.';
    return 0;
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << '.';
  cout << endl;
  int a = (n - k) / 2;
  for (int i = 1; i <= a; i++) cout << '.';
  for (int i = 1; i <= k; i++) cout << '#';
  for (int i = 1; i <= a; i++) cout << '.';
  cout << endl;
  for (int i = 1; i <= n; i++) cout << '.';
  cout << endl;
  for (int i = 1; i <= n; i++) cout << '.';
  return 0;
}
