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 $= 1$.
[samples]