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;
constexpr int MAXN = (1 << 20) + 5;
constexpr int MOD = 1e9 + 7;
constexpr long long MOD2 = 1e9 + 9;
constexpr double eps = 1e-6;
constexpr int INF = (1 << 30);
int dir[][2] = {{-1, 0},
                {0, 1},
                {1, 0},
                {0, -1},
                {
                    -1,
                    -1,
                },
                {1, 1},
                {1, -1},
                {-1, 1}};
const int N = 1e6 + 9;
char mp[4][105];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  int n, m;
  while (cin >> n >> m) {
    cout << "YES\n";
    for (int i = 0; i < 4; i++)
      for (int j = 0; j < 101; j++) mp[i][j] = '.';
    if (m % 2) {
      if (m <= 3) {
        mp[1][n / 2] = '#';
        if (m == 3) mp[1][n / 2 - 1] = '#', mp[1][n / 2 + 1] = '#';
      } else {
        mp[1][1] = '#', mp[2][1] = '#', mp[1][2] = '#';
        for (int i = 1; i < m / 2; i++) mp[1][2 + i] = '#', mp[2][2 + i] = '#';
      }
    } else {
      for (int i = 1; i <= m / 2; i++) mp[1][i] = '#', mp[2][i] = '#';
    }
    for (int i = 0; i < 4; i++) {
      for (int j = 0; j < n; j++) cout << mp[i][j];
      cout << '\n';
    }
  }
  return 0;
}
