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;
char t[200][200];
int main() {
  int a, b;
  cin >> a >> b;
  if (b <= (a - 2) * 2) {
    printf("YES\n");
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < a; j++) {
        t[i][j] = '.';
      }
    }
    if (b % 2 == 1) {
      for (int i = 1; i < 3 && b != 0; i++) {
        if (i == 1) {
          for (int j = 0; j < a / 2 && b != 0; j++) {
            if (j == 0) {
              t[i][a / 2] = '#';
              b--;
            } else {
              t[i][a / 2 - j] = '#';
              t[i][a / 2 + j] = '#';
              b -= 2;
            }
          }
        } else {
          for (int j = a / 2 - 1; j > 0 && b != 0; j--) {
            {
              t[i][a / 2 - j] = '#';
              t[i][a / 2 + j] = '#';
              b -= 2;
            }
          }
        }
      }
    } else {
      for (int i = 1; i < 3; i++) {
        for (int j = 1; j <= b / 2; j++) t[i][j] = '#';
      }
    }
    printf("%s\n%s\n%s\n%s\n", t[0], t[1], t[2], t[3]);
  } else {
    printf("NO");
  }
}
