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;
const long long INF = 0x3f3f3f3f;
const int N = 3e5 + 10;
const int M = 3e6 + 10;
const long long mod = 998244353;
const double PI = acos(-1);
inline long long ab(long long x) { return x < 0 ? -x : x; }
inline int mm(int x, int p) { return x >= p ? x - p : x < 0 ? x + p : x; }
bool ans[4][121];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  puts("YES");
  memset(ans, false, sizeof ans);
  for (int i = 1; i <= 2 && k >= 2; ++i) {
    for (int j = 1; j < (n >> 1) && k >= 2; ++j) {
      k -= 2;
      ans[i][j] = ans[i][n - 1 - j] = true;
    }
  }
  for (int i = 1; i <= 2; ++i)
    if (k) k--, ans[i][n >> 1] = true;
  for (int i = 0; i <= 3; ++i) {
    for (int j = 0; j < n; ++j)
      if (ans[i][j])
        putchar('#');
      else
        putchar('.');
    puts("");
  }
  return 0;
}
