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;
void pr1l(int n) {
  while (n--) cout << ".";
  cout << endl;
}
void pr1lh(int n, int k) {
  cout << ".";
  n--;
  while (n--) {
    if (k > 0) {
      cout << "#";
      k--;
    } else
      cout << ".";
  }
  cout << endl;
}
int main() {
  int n, k;
  cin >> n >> k;
  if (k % 2 == 0) {
    if (k / 2 <= n - 2) {
      cout << "YES" << endl;
      pr1l(n);
      int ctr = 0;
      for (int i = 0; i < n; i++) {
        if (i >= 1 && ctr < k / 2) {
          cout << "#";
          ctr++;
        } else
          cout << ".";
      }
      cout << endl;
      ctr = 0;
      for (int i = 0; i < n; i++) {
        if (i >= 1 && ctr < k / 2) {
          cout << "#";
          ctr++;
        } else
          cout << ".";
      }
      cout << endl;
      pr1l(n);
    } else
      cout << "NO" << endl;
  } else if (k <= n - 2) {
    cout << "YES" << endl;
    int offset = (n - k) / 2;
    pr1l(n);
    int ctr = 0;
    for (int i = 0; i < n; i++) {
      if (i >= offset && ctr < k) {
        cout << "#";
        ctr++;
      } else
        cout << ".";
    }
    cout << endl;
    pr1l(n);
    pr1l(n);
  } else if (k > n - 2 && k <= 2 * n - 4) {
    cout << "YES" << endl;
    pr1l(n);
    int ctr = 0;
    for (int i = 0; i < n; i++) {
      if (i > 0 && ctr < k / 2 + 1) {
        ctr++;
        if (ctr == k / 2) {
          cout << ".";
          continue;
        } else
          cout << "#";
      } else
        cout << ".";
    }
    cout << endl;
    pr1lh(n, k / 2 + 1);
    pr1l(n);
  } else
    cout << "NO" << endl;
  return 0;
}
