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;
const long long N = 1e5 + 7, MAXN = 1e9 + 7;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.precision(18);
  long long n, k;
  cin >> n >> k;
  char a[4][n];
  for (long long i = 0; i < 4; i++)
    for (long long j = 0; j < n; j++) a[i][j] = '.';
  if (k % 2 == 0) {
    for (long long i = 1; i < 3; i++)
      for (long long j = 1; j < k / 2 + 1; j++) a[i][j] = '#';
    cout << "YES\n";
    for (long long i = 0; i < 4; i++) {
      for (long long j = 0; j < n; j++) cout << a[i][j];
      cout << endl;
    }
  } else if (k % 2 == 1) {
    long long sim = n / 2;
    if (k > n - 2) {
      for (long long i = 1; i < n - 1; i++) a[1][i] = '#';
      k -= (n - 2);
      for (long long i = sim - (k / 2); i <= sim + (k / 2); i++)
        if (i != sim) a[2][i] = '#';
    } else {
      for (long long i = sim - (k / 2); i <= sim + (k / 2); i++) a[1][i] = '#';
    }
    cout << "YES\n";
    for (long long i = 0; i < 4; i++) {
      for (long long j = 0; j < n; j++) cout << a[i][j];
      cout << endl;
    }
  } else {
    cout << "NO\n";
  }
}
