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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
void seriy() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
char res[4][100];
long long n, k;
void out() {
  for (long long i = 0; i < 4; i++) {
    for (long long j = 0; j < n; j++) {
      cout << res[i][j];
    }
    cout << endl;
  }
}
signed main() {
  seriy();
  cin >> n >> k;
  long long l = k;
  cout << "YES\n";
  for (long long i = 0; i < 4; i++) {
    for (long long j = 0; j < n; j++) {
      res[i][j] = '.';
      if (k == 2 * (n - 2) && (i == 1 || i == 2) && j > 0 && j < n - 1) {
        res[i][j] = '#';
      }
    }
  }
  if (k == 2 * (n - 2)) {
    out();
    return 0;
  }
  for (long long i = 1; i < n / 2; i++) {
    if (l <= 0) {
      out();
      return 0;
    }
    if (l == 1) {
      res[1][n / 2] = '#';
      out();
      return 0;
    }
    res[1][i] = '#';
    res[1][n - i - 1] = '#';
    l -= 2;
  }
  for (long long i = 1; i < n / 2; i++) {
    if (l <= 0) {
      out();
      return 0;
    }
    if (l == 1) {
      res[1][n / 2] = '#';
      out();
      return 0;
    }
    res[2][i] = '#';
    res[2][n - i - 1] = '#';
    l -= 2;
  }
  if (l <= 0) {
    out();
    return 0;
  }
  if (l == 1) {
    res[1][n / 2] = '#';
    out();
    return 0;
  }
  return 0;
}
