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;
char a[5][110];
int n, k;
int main() {
  scanf("%d %d", &n, &k);
  if (k % 2 == 0) {
    printf("YES\n");
    k /= 2;
    for (int i = 2; i < 2 + k; i++) a[2][i] = a[3][i] = '#';
    for (int i = 1; i <= 4; i++, puts(""))
      for (int j = 1; j <= n; j++) printf("%c", (a[i][j] == '#' ? '#' : '.'));
  } else if (k % 2 == 1) {
    printf("YES\n");
    if (k <= n - 2) {
      k = (k - 1) / 2;
      for (int i = 0; i <= k; i++)
        a[2][(n + 1) / 2 + i] = a[2][(n + 1) / 2 - i] = '#';
    } else {
      for (int i = 2; i <= n - 1; i++) a[2][i] = '\#';
      k -= (n - 2);
      k /= 2;
      for (int i = 1; i <= k; i++) a[3][i + 1] = a[3][n - i] = '#';
    }
    for (int i = 1; i <= 4; i++, puts(""))
      for (int j = 1; j <= n; j++) printf("%c", (a[i][j] == '#' ? '#' : '.'));
  }
}
