Sum of Shortest Path Distances in the Sky Garden Maze

Prof. Du and Prof. Pang plan to build a sky garden near Allin city. The garden will include a plant maze formed by a set of circular and straight roads. In the blueprint, there are n circles (each circle i has radius \(i\)) centered at the origin \((0,0)\), and m straight lines (each passing through \((0,0)\)). Each circle is divided into \(2m\) equal arcs by these lines, and every intersection between a circle and a line gives two distinct points. In addition, any two different straight roads meet at the origin.

Let \(Q\) be the set of all \(n+m\) roads and let \(P\) be the set of all distinct intersections of any two different roads in \(Q\). Thus, \(|P|=2nm+1\) (the extra one corresponds to the origin).

For any two different points \(a, b\) in \(P\), define \(dis(\{a,b\})\) as the shortest distance to walk from \(a\) to \(b\) along the roads. Note that on a circle of radius \(i\), the distance along a circular arc between consecutive intersection points is \(\frac{\pi i}{m}\). On any straight road (which contains the intersections with the circles and the origin), the distance between the intersection points corresponding to circles with radii \(r\) and \(r+1\) is exactly 1.

Any point (other than the origin) can be represented in polar coordinates as \((r, \theta)\) where \(r\in\{1,2,\dots,n\}\) and \(\theta\) is one of the \(2m\) equally spaced angles: \(0,\,\frac{\pi}{m},\,\frac{2\pi}{m},\,\dots,\,\frac{(2m-1)\pi}{m}\). For two such points \((r_1,\theta_1)\) and \((r_2,\theta_2)\) (with \(r_1,r_2>0\)), one may travel radially and along a circular arc. In fact, if we choose an intermediate circle of radius \(t\) (with \(0\le t\le \min(r_1,r_2)\)), the travel distance is \[ r_1+r_2-2t+t\cdot\alpha, \quad\text{where } \alpha=\frac{\pi}{m}\cdot L, \] with \(L=\min(|k_1-k_2|,2m-|k_1-k_2|)\) when \(\theta_1=k_1\frac{\pi}{m}\) and \(\theta_2=k_2\frac{\pi}{m}\). Optimizing over \(t\) shows that when \(\frac{\pi L}{m}2\) it is optimal to choose \(t=0\) so that the distance becomes \(r_1+r_2\). (For points on the same ray, \(L=0\), and the distance is simply \(|r_1-r_2|\).)

Your task is to calculate the sum of \(dis(\{a,b\})\) for every unordered pair \(\{a,b\}\subseteq P\).

Input Format: Two integers \(n\) and \(m\).

Output Format: Output one real number representing the sum of shortest distances. Your answer will be accepted if it has an absolute or relative error of at most \(10^{-6}\).

inputFormat

The input consists of a single line with two integers \(n\) and \(m\) separated by a space.

outputFormat

Output a single real number: the sum of \(dis(\{a,b\})\) for all unordered pairs \(\{a,b\}\subseteq P\). The answer should be printed with at least 6 digits after the decimal point.

sample

1 1

4.000000