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;
using ll = long long;
using vi = vector<long long>;
using pii = pair<long long, long long>;
const signed INF_ = 1001001001;
const long long INF = 1001001001001001001LL;
const long long DX[9] = {0, 1, 0, -1, 1, 1, -1, -1, 0},
                DY[9] = {-1, 0, 1, 0, -1, 1, 1, -1, 0};
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (auto i = (begin(v)); i != (end(v)); ++i)
    os << *i << (i == end(v) - 1 ? "" : " ");
  return os;
}
template <class T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto i = (begin(v)); i != (end(v)); ++i) is >> *i;
  return is;
}
template <class T, class U>
istream &operator>>(istream &is, pair<T, U> &p) {
  is >> p.first >> p.second;
  return is;
}
template <class T, class U>
bool chmax(T &a, const U &b) {
  return a < b ? a = b, 1 : 0;
}
template <class T, class U>
bool chmin(T &a, const U &b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
using heap = priority_queue<T, vector<T>, greater<T>>;
struct {
  template <class T>
  operator T() {
    T x;
    cin >> x;
    return x;
  }
} IN;
struct before_main_function {
  before_main_function() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(15) << fixed;
  }
} before_main_function;
signed main() {
  long long n, k;
  cin >> n >> k;
  vector<string> ans(4, string(n, '.'));
  long long y = 1;
  if (n - 2 < k) {
    for (long long i = 0, ilen = (long long)(n - 2); i < ilen; ++i)
      ans[y][1 + i] = '#';
    k -= n - 2;
    y++;
  }
  if (k % 2) {
    ans[y][n / 2] = '#';
    k--;
  }
  for (long long i = 1, ilen = (long long)(k / 2 + 1); i < ilen; ++i) {
    ans[y][n / 2 - i] = ans[y][n / 2 + i] = '#';
  }
  cout << "YES"
       << "\n";
  for (long long i = 0, ilen = (long long)(4); i < ilen; ++i)
    cout << ans[i] << "\n";
  return 0;
}
