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 main() {
  int n, k;
  while (~scanf("%d%d", &n, &k)) {
    cout << "YES" << endl;
    if (k % 2 == 0) {
      int cnt = k / 2;
      for (int i = 1; i <= 4; i++) {
        for (int j = 1; j <= n; j++) {
          if (i == 1 || i == 4) {
            cout << ".";
          } else {
            if (j == 1 || j == n) {
              cout << ".";
            } else {
              if (j <= cnt + 1) {
                cout << "#";
              } else {
                cout << ".";
              }
            }
          }
        }
        cout << endl;
      }
    } else {
      if (k <= (n - 2)) {
        int cnt = (n - k) / 2;
        for (int i = 1; i <= 4; i++) {
          for (int j = 1; j <= n; j++) {
            if (i == 1 || i == 3 || i == 4) {
              cout << ".";
            } else {
              if (j == 1 || j == n)
                cout << ".";
              else {
                if (j > cnt && j < (cnt + k + 1))
                  cout << "#";
                else
                  cout << ".";
              }
            }
          }
          cout << endl;
        }
      } else {
        char a[5][105];
        for (int i = 1; i <= 4; i++) {
          for (int j = 1; j <= n; j++) a[i][j] = '.';
        }
        a[2][2] = a[3][2] = a[3][3] = '#';
        int cnt = k / 2 - 1;
        for (int i = 4; i < 4 + cnt; i++) a[2][i] = a[3][i] = '#';
        for (int i = 1; i <= 4; i++) {
          for (int j = 1; j <= n; j++) {
            cout << a[i][j];
          }
          cout << endl;
        }
      }
    }
  }
  return 0;
}
