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;
double minf(double a, double b) {
  if (a < b)
    return a;
  else
    return b;
}
int diff(int a, int b) { return abs(a - b); }
long long mul(long long a, long long b) { return (a * b) % 998244353; }
long long gcdf(long long x, long long y) {
  if (y == 0)
    return x;
  else
    return gcdf(y, x % y);
}
double power(double x) { return x * x; }
void solve() {
  int n, k;
  cin >> n >> k;
  vector<vector<char> > v(4, vector<char>(n, '.'));
  int idx1 = 1, idx2 = n - 2;
  while (k > 1 && idx1 < idx2) {
    v[1][idx1] = '#';
    v[1][idx2] = '#';
    idx1++;
    idx2--;
    k -= 2;
  }
  idx1 = 1;
  idx2 = n - 2;
  while (k > 1 && idx1 < idx2) {
    v[2][idx1] = '#';
    v[2][idx2] = '#';
    idx1++;
    idx2--;
    k -= 2;
  }
  idx1 = 1;
  while (k > 0) {
    v[idx1][(n - 1) / 2] = '#';
    k--;
    idx1++;
  }
  cout << "YES" << '\n';
  for (int i = 0; i < 4; i += 1) {
    for (int j = 0; j < n; j += 1) cout << v[i][j];
    cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
    cout << '\n';
  }
}
