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 & 1) && K / (N - 2)) {
    puts("YES");
    for (int i = 0; i < N; i++) {
      putchar('.');
    }
    puts("");
    putchar('.');
    for (int i = 1; i < N - 1; i++) {
      putchar('#');
    }
    putchar('.');
    puts("");
    putchar('.');
    for (int i = 1; i <= (K % (N - 2)) / 2; i++) {
      putchar('#');
    }
    for (int i = (K % (N - 2)) / 2 + 1; i < N - 1 - (K % (N - 2)) / 2; i++) {
      putchar('.');
    }
    for (int i = 1; i <= (K % (N - 2)) / 2; i++) {
      putchar('#');
    }
    putchar('.');
    puts("");
    for (int i = 0; i < N; i++) {
      putchar('.');
    }
    puts("");
  } else if (K & 1) {
    puts("YES");
    for (int i = 0; i < N; i++) {
      putchar('.');
    }
    puts("");
    for (int i = 0; i < (N - K) / 2; i++) {
      putchar('.');
    }
    for (int i = (N - K) / 2; i < N - (N - K) / 2; i++) {
      putchar('#');
    }
    for (int i = N - (N - K) / 2; i < N; i++) {
      putchar('.');
    }
    puts("");
    if (K >= N - 2) {
      putchar('.');
      for (int i = 1; i < N - 1; i++) {
        putchar('#');
      }
      putchar('.');
    } else {
      for (int i = 0; i < N; i++) {
        putchar('.');
      }
    }
    puts("");
    for (int i = 0; i < N; i++) {
      putchar('.');
    }
    puts("");
  } else {
    puts("YES");
    for (int i = 0; i < N; i++) {
      putchar('.');
    }
    puts("");
    putchar('.');
    for (int i = 1; i <= K / 2; i++) {
      putchar('#');
    }
    for (int i = K / 2 + 1; i < N; i++) {
      putchar('.');
    }
    puts("");
    putchar('.');
    for (int i = 1; i <= K / 2; i++) {
      putchar('#');
    }
    for (int i = K / 2 + 1; i < N; i++) {
      putchar('.');
    }
    puts("");
    for (int i = 0; i < N; i++) {
      putchar('.');
    }
    puts("");
  }
  return 0;
}
