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;
void _dbg(istream_iterator<string> it) { ++it; }
template <typename T, typename... Args>
void _dbg(istream_iterator<string> it, T a, Args... args) {
  if (*it == "'\\n'" || *it == "\"\\n\"" || *it == "endl") {
    cout << "\n";
  } else {
    cout << setw(8) << std::left << *it << " = " << setw(4) << std::right << a
         << "\n";
  }
  _dbg(++it, args...);
}
template <typename T>
std::ostream &_containerprint(std::ostream &out, T const &val) {
  return (out << setw(4) << val << " ");
}
template <typename T1, typename T2>
std::ostream &_containerprint(std::ostream &out, std::pair<T1, T2> const &val) {
  return (out << "{" << setw(4) << val.first << " " << setw(4) << val.second
              << "} ");
}
template <template <typename, typename...> class TT, typename... Args>
std::ostream &operator<<(std::ostream &out, const TT<Args...> &cont) {
  for (auto &&elem : cont) {
    _containerprint(out, elem);
  }
  return out;
}
std::ostream &operator<<(std::ostream &out, const std::string &s) {
  return operator<<<char>(out, s);
}
const double EPS = 1e-9;
const long long mod = 998244353;
template <typename T>
T gcd(T a, T b) {
  T c;
  while (b) {
    c = b;
    b = a % b;
    a = c;
  }
  return a;
}
template <typename T>
T powmod(T a, T b) {
  T res = 1;
  a %= mod;
  while (b) {
    if (b & 1) res = res * (a % mod);
    a = a * (a % mod);
    b >>= 1;
  }
  return res;
}
void Solve() {
  int n, k;
  cin >> n >> k;
  bool grid[4][101] = {};
  int x = 1, y = 1;
  if (k == 2 * (n - 2)) {
    for (y = 1; y <= 2; ++y)
      for (x = 1; x <= n - 2; ++x) grid[y][x] = 1;
  } else {
    while (k > 1) {
      if (x == n / 2) {
        x = 1;
        ++y;
      } else {
        grid[y][x] = grid[y][n - x - 1] = 1;
        k -= 2;
        ++x;
      }
    }
    if (k == 1) {
      grid[1][n / 2] = 1;
    }
  }
  cout << "YES\n";
  for (y = 0; y < 4; ++y) {
    for (x = 0; x < n; ++x) {
      if (grid[y][x])
        cout << "#";
      else
        cout << ".";
    }
    cout << '\n';
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  Solve();
}
