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 a[1000];
int main() {
  int n, k;
  cin >> n >> k;
  n -= 2;
  int cnt = 0;
  int t = 0;
  t = k / n + 1;
  cnt = k % n;
  if (cnt == 0 && t > 0) t--;
  if (k > n) {
    if (t > 2) {
      cout << "NO" << endl;
      return 0;
    }
  }
  n += 2;
  cout << "YES" << endl;
  if (k % 2) {
    t = 1;
    if (k > n - 2)
      cnt = k % (n - 2);
    else
      cnt = 0;
    k = min(k, n - 2);
  } else {
    t = 2;
    cnt = k / 2;
    k /= 2;
  }
  int temp1 = (n - k) / 2;
  int temp2 = 0;
  int temp3 = 0;
  if (cnt % 2)
    temp2 = (n - cnt) / 2;
  else {
    temp2 = (n - 1 - cnt) / 2;
    temp3 = n - temp2 * 2 - cnt;
  }
  for (int i = 0; i < n; i++) cout << ".";
  cout << endl;
  for (int i = 0; i < temp1; i++) cout << ".";
  for (int i = temp1, j = 1; i < n; i++, j++) {
    if (j <= k)
      cout << "#";
    else
      cout << ".";
  }
  cout << endl;
  for (int i = 0; i < temp2; i++) cout << ".";
  for (int i = temp2, j = 1; i < n; i++, j++) {
    if (cnt % 2 == 0 && t == 1) {
      if (j <= cnt / 2)
        cout << "#";
      else {
        cout << ".";
        temp3--;
        if (temp3 == 0) j = 0;
      }
    } else {
      if (j <= cnt)
        cout << "#";
      else
        cout << ".";
    }
  }
  cout << endl;
  for (int i = 0; i < n; i++) cout << ".";
  cout << endl;
}
