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 i, j, k, n, m, a, b, x, y, rest = 0;
  scanf("%d %d", &n, &k);
  string str[n + 5], temp;
  for (i = 0; i < n; i++) {
    temp.push_back('.');
  }
  printf("YES\n");
  for (i = 0; i < 4; i++) {
    str[i] = temp;
  }
  if (n == 3) {
    rest = k;
    y = n / 2;
    x = 1;
    for (i = 0; i < rest; i++) {
      str[x][y] = '#';
      x++;
    }
    for (i = 0; i < 4; i++) {
      cout << str[i] << endl;
    }
    return 0;
  }
  a = k / 2;
  int p = n / 2;
  b = a;
  x = 1;
  y = 1;
  while (b--) {
    if (y == p) {
      x++;
      if (x == 3) break;
      y = 1;
    }
    if (x == 3) break;
    str[x][y] = '#';
    y++;
  }
  rest += b + 1;
  b = a;
  x = 1;
  y = n - 2;
  while (b--) {
    if (y == p) {
      x++;
      if (x == 3) break;
      y = n - 2;
    }
    if (x == 3) break;
    str[x][y] = '#';
    y--;
  }
  rest += b + 1;
  if (k % 2) rest++;
  y = p;
  x = 1;
  for (i = 0; i < rest; i++) {
    str[x][y] = '#';
    x++;
  }
  for (i = 0; i < 4; i++) {
    cout << str[i] << endl;
  }
  return 0;
}
