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 EPS = 1e-15;
int toInt(string s) {
  int sm;
  stringstream second(s);
  second >> sm;
  return sm;
}
int dcmp(double x, double y) { return fabs(x - y) <= EPS ? 0 : x < y ? -1 : 1; }
int main() {
  int n, k, remaining = 0, flag = 0;
  std::cin >> n >> k;
  char grid[5][n + 1];
  for (int i = 0; i <= 4; ++i) {
    for (int j = 0; j < n + 1; ++j) {
      grid[i][j] = '.';
    }
  }
  if (k % 2) remaining = 1, k--;
  if (k % 2 == 0) {
    for (int i = 2; i <= n / 2; ++i) {
      if (k == 0) break;
      grid[2][i] = grid[2][n - i + 1] = '#';
      k -= 2;
    }
  }
  if (k % 2 == 0) {
    for (int i = 2; i <= n / 2; ++i) {
      if (k == 0) break;
      grid[3][i] = grid[3][n - i + 1] = '#';
      k -= 2;
    }
  }
  if (k % 2 == 0 && k > 0) {
    grid[2][n / 2 + 1] = grid[3][n / 2 + 1] = '#';
    k -= 2;
  }
  if (remaining) grid[2][n / 2 + 1] = '#';
  cout << "YES\n";
  for (int i = 1; i <= 4; ++i) {
    for (int j = 1; j < n + 1; ++j) {
      cout << grid[i][j];
    }
    cout << "\n";
  }
  return 0;
}
