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 double eps = 1e-12;
const double PI = acos(-1);
const long long INFF = (long long)1e18;
const int INF = (int)1e9;
const int mod = (int)1e9 + 7;
const int MAX = (int)5e2 + 7;
char st[MAX][MAX];
int main(void) {
  int n, k;
  scanf("%d %d", &n, &k);
  ;
  for (int i = (1); i < (5); ++i)
    for (int j = (1); j < (n + 1); ++j) st[i][j] = '.';
  if (k % 2) {
    st[2][n / 2 + 1] = '#';
    k--;
    for (int i = (1); i < (n / 2); ++i) {
      if (!k) break;
      st[2][n / 2 + 1 + i] = st[2][n / 2 + 1 - i] = '#';
      k -= 2;
    }
    for (int i = (1); i < (n / 2); ++i) {
      if (!k) break;
      st[3][1 + i] = st[3][n - i] = '#';
      k -= 2;
    }
  } else {
    for (int i = (2); i < (n + 1); ++i) {
      if (!k) break;
      st[2][i] = st[3][i] = '#';
      k -= 2;
    }
  }
  puts("YES");
  for (int i = (1); i < (5); ++i) printf("%s\n", &st[i][1]);
  return 0;
}
