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;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  if (k % 2 == 0) {
    printf("YES\n");
    for (int i = 0; i < 4; i++) {
      int cnt1 = k / 2;
      int cnt2 = k / 2;
      for (int j = 0; j < n; j++) {
        if (i == 1 && j > 0) {
          if (cnt1 > 0) {
            printf("#");
            cnt1--;
          } else
            printf(".");
        } else if (i == 2 && j > 0) {
          if (cnt2 > 0) {
            printf("#");
            cnt2--;
          } else
            printf(".");
        } else {
          printf(".");
        }
      }
      printf("\n");
    }
  } else if (k % 2 != 0 && k <= 2 * n - 5) {
    printf("YES\n");
    for (int i = 0; i < 4; i++) {
      if (k <= n - 2) {
        int cnt1 = k;
        for (int j = 0; j < n; j++) {
          if (i == 1 && j >= (n - k) / 2) {
            if (cnt1 > 0) {
              printf("#");
              cnt1--;
            } else {
              printf(".");
            }
          } else {
            printf(".");
          }
        }
      } else {
        int cnt1 = (k - n + 2) / 2;
        for (int j = 0; j < n; j++) {
          if (i == 1) {
            if (j != 0 && j != n - 1) {
              printf("#");
            } else {
              printf(".");
            }
          } else if (i == 2) {
            if (j >= 1 && j <= cnt1) {
              printf("#");
            } else if (j > n - 2 - cnt1 && j <= n - 2) {
              printf("#");
            } else
              printf(".");
          } else {
            printf(".");
          }
        }
      }
      printf("\n");
    }
  } else
    printf("NO\n");
  return 0;
}
