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;
const int M = 5e6 + 5;
const long long int mod = 1e9 + 7;
const long long int inf = 9e18;
const double pi = 3.14159265358979323846;
long long int t, n, m, x, sum, k, p, u = 0;
string s, z, ans = "";
long long int P[10000005] = {0};
void sieve() {
  for (long long int i = 2; i < (10000005); i++)
    if (!P[i])
      for (long long int j = i * i; j < (10000005); j += i) P[j] = i;
}
long long int raised(long long int x, long long int v) {
  if (v == 0) return 1 % mod;
  if (v % 2 == 1) return (x * raised(x, v - 1)) % mod;
  long long int r = raised(x, v / 2);
  return (r * r) % mod;
}
bool prime(long long int n) {
  if (n == 1) return 1;
  for (int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long int hcf(long long int a, long long int b) {
  if (a % b == 0) return b;
  return hcf(b, a % b);
}
bool comp(long long int a, long long int b) { return (a < b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> k;
  cout << "YES\n";
  string a[4];
  for (long long int i = 0; i < 4; i++) {
    string g(n, '.');
    a[i] = g;
  }
  for (long long int i = 1; i < n / 2; i++) {
    for (long long int j = 1; j <= 2; j++) {
      if (k < 2) break;
      a[j][i] = '#';
      a[j][n - 1 - i] = '#';
      k -= 2;
    }
    if (k < 2) break;
  }
  while (k--) {
    a[k + 1][n / 2] = '#';
  }
  for (long long int i = 0; i < 4; i++) cout << a[i] << "\n";
}
