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 maxn = 1e2 + 10;
const int maxm = 5e4 + 10;
const int INF = 0x3f3f3f3f;
int n, k;
int ans[maxn][maxn];
void Print() {
  for (int i = 0; i < n; i++) printf(".");
  printf("\n");
}
int main() {
  while (scanf("%d%d", &n, &k) != EOF) {
    if (k % 2 == 0) {
      printf("YES\n");
      Print();
      printf(".");
      for (int i = 1; i <= k / 2; i++) printf("#");
      for (int i = 1; i <= n - 1 - k / 2; i++) printf(".");
      printf("\n");
      printf(".");
      for (int i = 1; i <= k / 2; i++) printf("#");
      for (int i = 1; i <= n - 1 - k / 2; i++) printf(".");
      printf("\n");
      Print();
    } else {
      printf("YES\n");
      int lest = k;
      k--;
      Print();
      printf(".");
      int l = n / 2 + 1 - k / 2, r = n / 2 + 1 + k / 2;
      for (int i = 2; i <= n - 1; i++) {
        if (i >= l && i <= r) {
          if (lest) {
            printf("#");
            lest--;
          } else
            printf(".");
        } else
          printf(".");
      }
      printf(".\n");
      if (lest) {
        l = n / 2 + 1 - lest / 2, r = n / 2 + 1 + lest / 2;
        printf(".");
        for (int i = 2; i <= n - 1; i++) {
          if (i >= l && i <= r && i != n / 2 + 1) {
            if (lest) {
              printf("#");
              lest--;
            } else
              printf(".");
          } else
            printf(".");
        }
        printf(".\n");
      } else
        Print();
      Print();
    }
  }
  return 0;
}
