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(int argc, char *argv[]) {
  int n, h;
  cin >> n;
  cin >> h;
  cout << "YES" << endl;
  for (int i = 0; i < n; ++i) {
    cout << ".";
  }
  cout << endl;
  vector<char> x;
  x.push_back('.');
  for (size_t i = 1; i < n / 2; ++i) {
    if (h >= 2) {
      x.push_back('#');
      h -= 2;
    } else {
      x.push_back('.');
    }
  }
  for (auto y = x.begin(); y != x.end(); ++y) {
    cout << *y;
  }
  if (h > 0) {
    cout << "#";
    h--;
  } else {
    cout << ".";
  }
  for (auto y = x.rbegin(); y != x.rend(); ++y) {
    cout << *y;
  }
  cout << endl;
  vector<char> z;
  z.push_back('.');
  for (size_t i = 1; i < n / 2; ++i) {
    if (h >= 2) {
      z.push_back('#');
      h -= 2;
    } else {
      z.push_back('.');
    }
  }
  for (auto y = z.begin(); y != z.end(); ++y) {
    cout << *y;
  }
  if (h > 0) {
    cout << "#";
    h--;
  } else {
    cout << ".";
  }
  for (auto y = z.rbegin(); y != z.rend(); ++y) {
    cout << *y;
  }
  cout << endl;
  for (int i = 0; i < n; ++i) {
    cout << ".";
  }
  cout << endl;
  return 0;
}
