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;
long long add(long long x, long long y) {
  x += y;
  while (x >= 998244353) x -= 998244353;
  while (x < 0) x += 998244353;
  return x;
}
long long mul(long long x, long long y) { return (x * 1ll * y) % 998244353; }
long long binpow(long long x, long long y) {
  long long z = 1;
  while (y > 0) {
    if (y % 2 == 1) z = mul(z, x);
    x = mul(x, x);
    y /= 2;
  }
  return z;
}
long long inv(long long x) { return binpow(x, 998244353 - 2); }
long long divide(long long x, long long y) { return mul(x, inv(y)); }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k;
  cin >> n >> k;
  string str[4];
  for (long long i = 0; i < (long long)4; ++i) str[i] = string(n, '.');
  for (long long i = 1; i < n / 2 && k >= 2; i++) {
    str[1][i] = '#';
    str[1][n - 1 - i] = '#';
    k -= 2;
  }
  for (long long i = 1; i < n / 2 && k >= 2; i++) {
    str[2][i] = '#';
    str[2][n - 1 - i] = '#';
    k -= 2;
  }
  if (k == 1)
    str[1][n / 2] = '#';
  else if (k == 2)
    str[1][n / 2] = str[2][n / 2] = '#';
  cout << "YES\n";
  for (long long i = 0; i < (long long)4; ++i) cout << str[i] << endl;
  return 0;
}
