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][200];
int main(void) {
  int i, j, n, m;
  scanf("%d%d", &n, &m);
  memset(a, 0, sizeof(a));
  for (i = 1; i <= 4; i++)
    for (j = 1; j <= n; j++) a[i][j] = '.';
  if (m % 2 == 0) {
    for (j = 2; j < n && m >= 2; j++) {
      a[2][j] = a[3][j] = '#';
      m -= 2;
    }
    printf("YES\n");
    printf("%s\n%s\n%s\n%s", &a[1][1], &a[2][1], &a[3][1], &a[4][1]);
  } else if (m >= 5) {
    a[2][2] = a[3][2] = a[2][3] = '#';
    m -= 3;
    for (j = 4; j < n && m; j++) {
      a[2][j] = a[3][j] = '#';
      m -= 2;
    }
    printf("YES\n");
    printf("%s\n%s\n%s\n%s", &a[1][1], &a[2][1], &a[3][1], &a[4][1]);
  } else if (m == 1) {
    a[2][(n + 1) / 2] = '#';
    printf("YES\n");
    printf("%s\n%s\n%s\n%s", &a[1][1], &a[2][1], &a[3][1], &a[4][1]);
  } else if (m == 3) {
    int k = (n + 1) / 2;
    a[2][k] = a[2][k - 1] = a[2][k + 1] = '#';
    printf("YES\n");
    printf("%s\n%s\n%s\n%s", &a[1][1], &a[2][1], &a[3][1], &a[4][1]);
  }
  return 0;
}
