Balloon Connect-the-Dots

Have you ever played connect-the-dots? For those of us with no talent for drawing, it is one of the few ways to end up with a nice picture. The page has $n$ dots, each labelled with an integer between $1$ and $n$. You draw $n-1$ straight lines: one between $1$ and $2$, one between $2$ and $3$, one between $3$ and $4$, and so on to the end. Here is an example before and after connecting the dots.

Today you are drawing a connect-the-dots picture on the surface of a balloon. Assume the balloon is a perfect sphere. It starts with a radius of 1 centimetre. I have drawn $n$ dots on the surface and labelled them from $1$ to $n$. As the balloon inflates, the dots move further and further apart. My goal is the largest possible picture, so the total length of the $n-1$ line segments should be as large as possible.

Because I do not want the image to look faded, I will first inflate the balloon and only then draw the $n-1$ lines. I can increase the volume of the balloon by $v$ cubic centimetres per minute, and I can draw at a speed of $d$ centimetres per minute. When I draw the line from point $i$ to point $i+1$, I draw it along the surface of the balloon on a shortest path. Sometimes several shortest paths exist, but this problem only counts lengths, so it does not matter which one I draw.

I have $T$ minutes in total. The time spent inflating the balloon plus the time spent drawing must be at most $T$ minutes. When I stop, all $n-1$ lines must be fully drawn. Find the largest total line length I can draw.

The first line contains four integers $n$, $T$, $v$ and $d$, where $n$ ($2 \le n \le 1000$) is the number of points, $T$ ($1 \le T \le 10^6$) is the time available to both inflate the balloon and draw the $n-1$ lines in minutes, $v$ ($1 \le v \le 1000$) is the rate at which I can inflate the balloon in cubic centimetres per minute, and $d$ ($1 \le d \le 1000$) is the speed at which I can draw in centimetres per minute.

The next $n$ lines describe the points in order from $1$ to $n$, with the balloon centred at the origin. Each line holds two real numbers $x$ and $y$ and an integer $s$ ($s \in \{-1, 1\}$), giving the point on the surface at $(x, y, z)$ where $z = s\sqrt{1 - x^2 - y^2}$. It is guaranteed that $0 \le x^2 + y^2 \le 1$, and $x$ and $y$ are in centimetres. Each $x$ and $y$ is given to at most 6 digits after the decimal point.

Lines may cross one another or overlap completely, so a segment of the surface may be counted more than once. It is guaranteed that I can draw the picture on the balloon of radius one in at most $T$ minutes.

Print the length in centimetres of the largest completed drawing I can make. Round to 6 digits after the decimal point and print exactly 6 of them.

Be careful when $x^2 + y^2 = 1$. Rounding can leave $1 - x^2 - y^2$ stored as a tiny negative value such as $-0.0000000001$, and taking its square root then fails. For the same reason the cosine of the angle between two points can land slightly outside $[-1, 1]$. Clamping that value to the range before taking the arc cosine is the safe fix.