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, col;
  cin >> n >> k;
  if ((n - 2) * 2 < k) {
    printf("NO\n");
    return 0;
  }
  col = n - 2;
  if (k % 2 == 0) {
    printf("YES\n");
    for (int i = 1; i <= n; i++) {
      printf(".");
    }
    printf("\n");
    printf(".");
    int i = 1;
    for (i; i <= k / 2; i++) {
      printf("#");
    }
    for (i; i < n; i++) printf(".");
    printf("\n");
    printf(".");
    i = 1;
    for (i; i <= k / 2; i++) {
      printf("#");
    }
    for (i; i < n; i++) printf(".");
    printf("\n");
    for (int i = 1; i <= n; i++) {
      printf(".");
    }
    printf("\n");
  } else {
    if (k == col) {
      printf("YES\n");
      for (int i = 1; i <= n; i++) printf(".");
      printf("\n");
      printf(".");
      for (int i = 1; i <= k; i++) printf("#");
      printf(".");
      printf("\n");
      for (int i = 1; i <= n; i++) printf(".");
      printf("\n");
      for (int i = 1; i <= n; i++) printf(".");
      printf("\n");
    } else if (k < col) {
      printf("YES\n");
      for (int i = 1; i <= n; i++) printf(".");
      printf("\n");
      int ex = (col - k) / 2;
      ex++;
      for (int i = 1; i <= ex; i++) printf(".");
      for (int i = 1; i <= k; i++) printf("#");
      for (int i = 1; i <= ex; i++) printf(".");
      printf("\n");
      for (int i = 1; i <= n; i++) printf(".");
      printf("\n");
      for (int i = 1; i <= n; i++) printf(".");
      printf("\n");
    } else {
      printf("YES\n");
      for (int i = 1; i <= n; i++) printf(".");
      printf("\n");
      int ex = (k - col) / 2;
      printf(".");
      for (int i = 1; i <= col; i++) printf("#");
      printf(".\n");
      int i = 1;
      printf(".");
      for (int j = 1; j <= ex; j++) {
        printf("#");
        i++;
      }
      for (i; i <= col - ex; i++) printf(".");
      for (int i = 1; i <= ex; i++) printf("#");
      printf(".\n");
      for (int i = 1; i <= n; i++) printf(".");
      printf("\n");
    }
  }
  return 0;
}
