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;
using ll = long long;
using ull = unsigned long long;
template <typename... Args>
void writeln(Args... args) {
  ((cout << args << " "), ...);
  cout << endl;
}
template <typename... Args>
void _db(Args... args) {}
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  vector<char> v1(n, '.');
  vector<char> v2(n, '.');
  if (k % 2 == 0) {
    for (int i = 1; i <= k / 2; ++i) {
      v1[i] = '#';
      v2[i] = '#';
    }
  } else {
    if (k > n - 2) {
      for (int i = 1; i < n - 1; ++i) {
        v1[i] = '#';
      }
      int left = k - (n - 2);
      _db("left", left);
      int mid = n / 2;
      _db("mid is", mid);
      int inc = left / 2;
      _db("inc is", inc);
      for (int i = 0; i < inc; ++i) {
        _db("set left", mid - 1 - inc);
        _db("set right", mid + 1 + inc);
        v2[mid - 1 - i] = '#';
        v2[mid + 1 + i] = '#';
      }
    } else {
      for (int i = (n / 2) - (k / 2); i <= (n / 2) + (k / 2); ++i) {
        v1[i] = '#';
      }
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; ++i) {
    cout << '.';
  }
  cout << endl;
  for (char c : v1) {
    cout << c;
  }
  cout << endl;
  for (char c : v2) {
    cout << c;
  }
  cout << endl;
  for (int i = 0; i < n; ++i) {
    cout << '.';
  }
  cout << endl;
  return 0;
}
