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 int mod = 1000000000 + 7;
const long long int N = 10000000 + 6;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  char matrix[4][n];
  for (int i = 0; i < 4; i++)
    for (int j = 0; j < n; j++) matrix[i][j] = '.';
  bool bo = true;
  if (k & 1) {
    if (k > n - 2) {
      int g = n - 2;
      for (int i = 1; i <= g; i++) matrix[1][i] = '#';
      int l = k - n;
      matrix[2][1] = '#';
      matrix[2][n - 2] = '#';
      for (int i = 2; i <= l + 1; i++) matrix[2][i] = '#';
    } else {
      int i = n / 2 - k / 2;
      int c = 1;
      for (int j = i; c <= k; c++, j++) matrix[1][j] = '#';
    }
  } else {
    int g = k / 2;
    for (int i = 1; i <= g; i++) matrix[1][i] = '#';
    for (int i = 1; i <= g; i++) matrix[2][i] = '#';
  }
  if (bo) {
    cout << "YES\n";
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) cout << matrix[i][j];
      cout << '\n';
    }
  } else
    cout << "NO\n";
  return 0;
}
