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>
int main() {
  int n;
  int k;
  int x, y;
  int type;
  int odd_num;
  scanf("%d", &n);
  scanf("%d", &k);
  odd_num = ((n - 2) - k) / 2;
  if (k % 2 == 0) {
    printf("YES\n");
    type = 0;
  }
  if (k % 2 != 0 && k >= 5 && n >= 5) {
    printf("YES\n");
    type = 1;
  }
  if (k % 2 != 0 && k < 5 && n % 2 != 0) {
    printf("YES\n");
    type = 2;
  }
  if (k % 2 != 0 && k < 5 && n % 2 == 0) {
    printf("NO");
    exit(0);
  }
  for (x = 0; x < n; x++) {
    printf(".");
  }
  printf("\n");
  for (y = 1; y < 3; y++) {
    printf(".");
    if (type == 0) {
      for (x = 0; x < k / 2; x++) {
        printf("#");
      }
      for (x = 0; x < n - (k / 2 + 2); x++) {
        printf(".");
      }
    }
    if (type == 1 && y == 1) {
      for (x = 0; x < (k + 1) / 2; x++) {
        printf("#");
      }
      for (x = 0; x < n - (((k + 1) / 2) + 2); x++) {
        printf(".");
      }
    }
    if (type == 1 && y == 2) {
      printf("#");
      for (x = 0; x < (k - 5) / 2; x++) {
        printf("#");
      }
      printf(".");
      printf("#");
      for (x = 0; x < n - (((k + 1) / 2) + 2); x++) {
        printf(".");
      }
    }
    if (type == 2 && y == 1) {
      for (x = 0; x < odd_num; x++) {
        printf(".");
      }
      for (x = 0; x < k; x++) {
        printf("#");
      }
      for (x = 0; x < odd_num; x++) {
        printf(".");
      }
    }
    if (type == 2 && y == 2) {
      for (x = 0; x < n - 2; x++) {
        printf(".");
      }
    }
    printf(".");
    printf("\n");
  }
  for (x = 0; x < n; x++) {
    printf(".");
  }
  return (0);
}
