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;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 10;
char a[5][110];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  printf("YES\n");
  if (k % 2 == 0) {
    k /= 2;
    int x = k;
    for (int i = 0; i < 4; ++i) {
      x = k;
      for (int j = 0; j < n; ++j) {
        if (i > 0 && i < 3 && j > 0 && j < n - 1 && x > 0) {
          a[i][j] = '#';
          x--;
        } else {
          a[i][j] = '.';
        }
      }
    }
  } else {
    if (k <= n - 2) {
      int x = (n - 2 - k) / 2;
      int cnt = 0;
      for (int i = 0; i < 4; ++i) {
        for (int j = 0; j < n; ++j) {
          if (i == 1) {
            if (cnt >= x + 1 && cnt - x <= k) {
              a[i][j] = '#';
            } else {
              a[i][j] = '.';
            }
            cnt++;
          } else {
            a[i][j] = '.';
          }
        }
      }
    } else {
      k -= n - 2;
      for (int i = 0; i < 4; ++i) {
        for (int j = 0; j < n; ++j) {
          if (i == 1 && j > 0 && j < n - 1) {
            a[i][j] = '#';
          } else if (i == 2) {
            if (j > 0 && j < n - 1 && k > 0) {
              a[i][j] = '#';
              a[i][n - 1 - j] = '#';
              k -= 2;
            } else if (a[i][j] != '#') {
              a[i][j] = '.';
            }
          } else {
            a[i][j] = '.';
          }
        }
      }
    }
  }
  for (int i = 0; i < 4; ++i) {
    for (int j = 0; j < n; ++j) {
      printf("%c", a[i][j]);
    }
    printf("\n");
  }
  printf("\n");
  return 0;
}
