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>
int main() {
  int a = 0, b = 0, i = 0, j = 0;
  scanf("%d %d", &a, &b);
  if (b <= (a - 2) * 2) {
    printf("Yes\n");
    if (b % 2 == 0) {
      for (i = 1; i <= 4; i++) {
        if (i == 1 || i == 4) {
          for (j = 1; j <= a; j++) printf(".");
          printf("\n");
        } else {
          for (j = 1; j <= a; j++) {
            if (j > 1 && j <= 1 + b / 2)
              printf("#");
            else
              printf(".");
          }
          printf("\n");
        }
      }
    } else {
      for (i = 1; i <= 4; i++) {
        if (i == 1 || i == 4) {
          for (j = 1; j <= a; j++) printf(".");
          printf("\n");
        } else if (i == 2) {
          if (b <= a - 2) {
            for (j = 1; j <= (a - b) / 2; j++) printf(".");
            for (j = 1; j <= b; j++) printf("#");
            for (j = 1; j <= (a - b) / 2; j++) printf(".");
            printf("\n");
          } else {
            printf(".");
            for (j = 1; j <= a - 2; j++) printf("#");
            printf(".\n");
          }
        } else if (i == 3) {
          if (b <= a - 2) {
            for (j = 1; j <= a; j++) printf(".");
            printf("\n");
          } else {
            printf(".");
            for (j = 1; j <= (b - (a - 2)) / 2; j++) printf("#");
            for (j = 1; j <= 2 * a - 4 - b; j++) printf(".");
            for (j = 1; j <= (b - (a - 2)) / 2; j++) printf("#");
            printf(".\n");
          }
        }
      }
    }
  } else
    printf("No\n");
  return 0;
}
