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 n, k;
char a[6][110];
int main() {
  cin >> n >> k;
  if (k > (n - 2) * 2) {
    cout << "NO"
         << "\n";
    return 0;
  }
  cout << "YES"
       << "\n";
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= n; j++) {
      a[i][j] = '.';
    }
  }
  int t = k;
  if (n % 2 == 1) {
    if (k % 2 == 1) {
      for (int i = 2; i <= 3; i++) {
        if (t <= 1) break;
        for (int j = 2; j <= (k + 1) / 2; j++) {
          if (t <= 1) break;
          if (j >= (n + 1) / 2) break;
          a[i][j] = '#';
          a[i][n - j + 1] = '#';
          t -= 2;
        }
      }
    } else {
      if (t <= 2) {
        for (int i = 2; i <= 3; i++) {
          if (t <= 0) break;
          a[i][(n + 1) / 2] = '#';
          t -= 1;
        }
      } else {
        for (int i = 2; i <= 3; i++) {
          if (t <= 0) break;
          for (int j = 2; j < (n + 1) / 2; j++) {
            if (t <= 0) break;
            a[i][j] = '#';
            a[i][n - j + 1] = '#';
            t -= 2;
          }
        }
      }
    }
  } else {
    if (k % 2 == 1) {
      cout << "NO"
           << "\n";
      return 0;
    } else {
      for (int i = 2; i <= 3; i++) {
        if (t <= 0) break;
        for (int j = 2; j <= n / 2; j++) {
          if (t <= 0) break;
          a[i][j] = '#';
          a[i][n - j + 1] = '#';
          t -= 2;
        }
      }
    }
  }
  if (t > 0) {
    for (int i = 2; i <= 3; i++) {
      if (t <= 0) break;
      a[i][(n + 1) / 2] = '#';
      t -= 1;
    }
  }
  for (int i = 1; i <= 4; i++) {
    for (int j = 1; j <= n; j++) {
      cout << a[i][j];
    }
    cout << "\n";
  }
}
