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 a[4][99];
void doprint(int n) {
  for (int j = 0; j < 4; j++) {
    for (int i = 0; i < n; i++) cout << a[j][i];
    cout << endl;
  }
}
void print(int n, int k) {
  for (int i = 0; i < n; i++) a[0][i] = a[1][i] = a[2][i] = a[3][i] = '.';
  if (k == 0) {
    doprint(n);
    return;
  }
  int mid = (n - 1) / 2;
  if (k <= 2) {
    a[2][mid] = '#';
    if (k == 2) a[1][mid] = '#';
    doprint(n);
    return;
  }
  if (k % 2 == 1) {
    a[2][mid] = '#';
    k--;
  } else {
    a[2][mid] = a[1][mid] = '#';
    k -= 2;
  }
  int ptr = 1;
  int p = 2;
  int cnt = 1;
  while (k > 0) {
    a[p][mid + ptr] = a[p][mid - ptr] = '#';
    ptr++;
    cnt += 2;
    if (cnt >= (n - 2)) {
      p = 1;
      ptr = 1;
      cnt = 1;
    }
    k -= 2;
  }
  doprint(n);
  return;
}
int main() {
  int n, k;
  cin >> n >> k;
  cout << "YES" << endl;
  print(n, k);
  return 0;
}
