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>
long long int mod = 1000000007;
using namespace std;
void solve() {
  long long int n, k;
  cin >> n >> k;
  char ans[4][n];
  int nval = n - 2;
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) {
      ans[i][j] = '.';
    }
  }
  int mid = nval / 2;
  mid++;
  if (k % 2 == 1) {
    for (int i = 1; i < 3; i++) {
      if (k <= 0) break;
      for (int j = mid; j >= 1; j--) {
        if (i > 1 && k % 2 == 0 && k < nval) {
          ans[i][1] = '#';
          ans[i][n - 2] = '#';
          k--;
          k--;
        }
        if (k > 0) {
          ans[i][j] = '#';
          k--;
          if (k <= 0) break;
          ans[i][mid + (mid - j)] = '#';
          if (j != mid) k--;
        } else
          break;
      }
    }
  } else {
    for (int i = 1; i < n - 1; i++) {
      for (int j = 1; j < 3; j++) {
        if (k) {
          ans[j][i] = '#';
          k--;
        }
      }
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < 4; i++) {
    for (int j = 0; j < n; j++) {
      cout << ans[i][j];
    }
    cout << endl;
  }
}
int main() {
  solve();
  return 0;
}
