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;
int32_t main() {
  long long t = 1;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    cout << "YES" << endl;
    char arr[2][n];
    for (long long i = 0; i < 2; i++)
      for (long long j = 0; j < n; j++) arr[i][j] = '.';
    long long it1 = 1, it2 = 1, rw = 0;
    if (k % 2 == 0) {
      while (k) arr[1][it1] = '#', arr[0][it1++] = '#', k -= 2;
    } else if (k % 2) {
      arr[rw][n / 2] = '#';
      k--;
      while (k) {
        if (it2 == n / 2) rw++, it2 = 1;
        arr[rw][it2] = '#', arr[rw][n - (++it2)] = '#', k -= 2;
      }
    }
    for (long long i = 0; i < n; i++) cout << ".";
    cout << endl;
    for (long long i = 0; i < n; i++) cout << arr[0][i];
    cout << endl;
    for (long long i = 0; i < n; i++) cout << arr[1][i];
    cout << endl;
    for (long long i = 0; i < n; i++) cout << ".";
    cout << endl;
  }
  return 0;
}
