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;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  if (0) cout << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',')
    if (0) cout << *sdbg++;
  if (0) cout << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
const int N = 2;
bool grid[N + 2][107];
int32_t main() {
  int m, k;
  scanf("%d%d", &m, &k);
  puts("YES");
  int sr = (m + 1) / 2;
  int lewe = k / 2;
  int cnt = 0;
  for (int i = 1; i <= 2; ++i) {
    for (int j = 2; j < sr; ++j) {
      if (cnt < lewe) {
        ++cnt;
        --k;
        grid[i][j] = 1;
      }
    }
  }
  for (int i = 1; i <= 2; ++i) {
    for (int j = sr + 1; j < m; ++j) {
      grid[i][j] = grid[i][sr - (j - sr)];
      k -= grid[i][j];
    }
  }
  for (int i = 1; i <= 2; ++i)
    if (k) {
      grid[i][sr] = 1;
      --k;
    }
  for (int i = 0; i < 4; ++i) {
    for (int j = 1; j <= m; ++j) {
      if (grid[i][j])
        printf("#");
      else
        printf(".");
    }
    puts("");
  }
}
