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;
vector<string> s(4);
int n, k;
void outp() {
  cout << "YES";
  for (int i = 0; i < 4; i++) {
    cout << "\n";
    for (int j = 0; j < n; j++) cout << s[i][j];
  }
  exit(0);
}
void t1() {
  for (int i = 1; i <= k / 2; i++) {
    s[1][i] = '#';
    s[2][i] = '#';
  }
  outp();
}
void t2() {
  for (int i = 1; i < n - 1; i++) s[1][i] = '#';
  s[2][n - 2] = '#';
  for (int i = 1; i <= k - n + 1; i++) s[2][i] = '#';
  outp();
}
void t3() {
  int i, j, kk;
  i = n / 2;
  j = i;
  kk = k + 1;
  while (kk > 0) {
    s[1][i] = '#';
    s[1][j] = '#';
    i++;
    j--;
    kk -= 2;
  }
  outp();
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < 4; i++)
    for (int j = 0; j < n; j++) s[i] += '.';
  if (k <= 2 * n - 4) {
    if (k % 2 == 0) t1();
    if (k >= n) t2();
    if (k % 2 == 1 && k <= n - 2) t3();
  }
  cout << "NO";
}
