Max Score: $850$ Points

Problem statement

There is a circle which radius is 1.
There are $N$ vertices on the circle's circumference.
The vertices divides into $N$ equal parts over the circumference.

You can choose $3$ distinct vertices, and you can make a triangle.
There are $\frac{N(N - 1)(N - 2)}{6}$ ways choosing vertices. The question is: Calculate the area of $K$-th smallest triangle in $\frac{N(N-1)(N-2)}{6}$ triangles.
If the area is same, you can order in any order.

If $N = 4, K = 3$, the result is following:
  • If you select vertices $1$, $2$, and $3$, the area of triangle $= 1$.
  • If you select vertices $1$, $2$, and $4$, the area of triangle $= 1$.
  • If you select vertices $1$, $3$, and $4$, the area of triangle $= 1$.
  • If you select vertices $2$, $3$, and $4$, the area of triangle $= 1$.
As a result, the 3rd smallest triangle's area $= 1$.

Input

The input is given from Standard Input in the following format:
$N \ K$

Output

  • Please output the $K$-th triangle area.
  • Print a floating number denoting the answer. The relative or absolute error of your answer should not be higher than $10^{−9}$.

Constraints

  • $3 \le N \le 200,000$
  • $1 \le K \le \frac{N(N-1)(N-2)}{6}$

Subtasks

Subtask 1 [ $160$ points ]
  • $N \le 100$
Subtask 2 [ $240$ points ]
  • $N \le 1000$
Subtask 3 [ $450$ points ]
  • $N \le 200,000$

Sample Input 1

4 3

Sample Output 1

1.0000000000000
This example is already explained in the problem statement.

Sample Input 2

6 9

Sample Output 2

0.86602540378
There are $6$ ways to choose a triangle which area is $\frac{\sqrt{3}}{4}$.
There are $10$ ways to choose a triangle which area is $\frac{\sqrt{3}}{2}$.
There are $2$ ways to choose a triangle which area is $\frac{3 \sqrt{3}}{4}$.
Therefore, the 9th smallest triangle's area is $\frac{\sqrt{3}}{2}$.

Sample Input 3

12 220

Sample Output 3

1.29903810568
Writer: E869120