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;
  cin >> n >> k;
  cout << "YES" << endl;
  if (k % 2) {
    if (k > n - 2) {
      int two = k - n + 2;
      for (int i = 0; i < n; i++) printf(".");
      printf("\n.");
      for (int i = 0; i < n - 2; i++) printf("#");
      printf(".\n.");
      for (int i = 0; i < two / 2; i++) printf("#");
      for (int i = 0; i < (n - 2) / 2 - (two / 2); i++) printf(".");
      printf(".");
      for (int i = 0; i < (n - 2) / 2 - (two / 2); i++) printf(".");
      for (int i = 0; i < two / 2; i++) printf("#");
      printf(".\n");
      for (int i = 0; i < n; i++) printf(".");
      printf("\n");
    } else {
      int tt = (n - 2) / 2 - (k / 2);
      for (int i = 0; i < n; i++) printf(".");
      printf("\n.");
      for (int i = 0; i < k / 2; i++) printf("#");
      for (int i = 0; i < tt; i++) printf(".");
      printf("#");
      for (int i = 0; i < tt; i++) printf(".");
      for (int i = 0; i < k / 2; i++) printf("#");
      printf(".\n");
      for (int i = 0; i < n; i++) printf(".");
      printf("\n");
      for (int i = 0; i < n; i++) printf(".");
      printf("\n");
    }
  } else {
    for (int i = 0; i < n; i++) printf(".");
    printf("\n.");
    for (int i = 0; i < k / 2; i++) printf("#");
    for (int i = 0; i < (n - k / 2 - 1); i++) printf(".");
    printf("\n.");
    for (int i = 0; i < k / 2; i++) printf("#");
    for (int i = 0; i < (n - k / 2 - 1); i++) printf(".");
    printf("\n");
    for (int i = 0; i < n; i++) printf(".");
    printf("\n");
  }
}
