Company Acquisitions
Contest: Codeforces Round 505 (rated, Div. 1 + Div. 2, based on VK Cup 2018 Final)
Difficulty rating: 3200
Tags: constructive algorithms, math

There are $$$n$$$ startups. Startups can be active or acquired. If a startup is acquired, then that means it has exactly one active startup that it is following. An active startup can have arbitrarily many acquired startups that are following it. An active startup cannot follow any other startup.

The following steps happen until there is exactly one active startup. The following sequence of steps takes exactly 1 day.

1. Two distinct active startups $$$A$$$, $$$B$$$, are chosen uniformly at random.
2. A fair coin is flipped, and with equal probability, $$$A$$$ acquires $$$B$$$ or $$$B$$$ acquires $$$A$$$ (i.e. if $$$A$$$ acquires $$$B$$$, then that means $$$B$$$'s state changes from active to acquired, and its starts following $$$A$$$).
3. When a startup changes from active to acquired, all of its previously acquired startups become active.

For example, the following scenario can happen: Let's say $$$A$$$, $$$B$$$ are active startups. $$$C$$$, $$$D$$$, $$$E$$$ are acquired startups under $$$A$$$, and $$$F$$$, $$$G$$$ are acquired startups under $$$B$$$:

Active startups are shown in red.

If $$$A$$$ acquires $$$B$$$, then the state will be $$$A$$$, $$$F$$$, $$$G$$$ are active startups. $$$C$$$, $$$D$$$, $$$E$$$, $$$B$$$ are acquired startups under $$$A$$$. $$$F$$$ and $$$G$$$ have no acquired startups:

If instead, $$$B$$$ acquires $$$A$$$, then the state will be $$$B$$$, $$$C$$$, $$$D$$$, $$$E$$$ are active startups. $$$F$$$, $$$G$$$, $$$A$$$ are acquired startups under $$$B$$$. $$$C$$$, $$$D$$$, $$$E$$$ have no acquired startups:

You are given the initial state of the startups. For each startup, you are told if it is either acquired or active. If it is acquired, you are also given the index of the active startup that it is following.

You're now wondering, what is the expected number of days needed for this process to finish with exactly one active startup at the end.

It can be shown the expected number of days can be written as a rational number $$$P/Q$$$, where $$$P$$$ and $$$Q$$$ are co-prime integers, and $$$Q \not= 0 \pmod{10^9+7}$$$. Return the value of $$$P \cdot Q^{-1}$$$ modulo $$$10^9+7$$$.

Input format:
The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 500$$$), the number of startups.

The next line will contain $$$n$$$ space-separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$a_i = -1$$$ or $$$1 \leq a_i \leq n$$$). If $$$a_i = -1$$$, then that means startup $$$i$$$ is active. Otherwise, if $$$1 \leq a_i \leq n$$$, then startup $$$i$$$ is acquired, and it is currently following startup $$$a_i$$$. It is guaranteed if $$$a_i \not= -1$$$, then $$$a_{a_i} =-1$$$ (that is, all startups that are being followed are active).

Output format:
Print a single integer, the expected number of days needed for the process to end with exactly one active startup, modulo $$$10^9+7$$$.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first sample, there are three active startups labeled $$$1$$$, $$$2$$$ and $$$3$$$, and zero acquired startups. Here's an example of how one scenario can happen

1. Startup $$$1$$$ acquires startup $$$2$$$ (This state can be represented by the array $$$[-1, 1, -1]$$$)
2. Startup $$$3$$$ acquires startup $$$1$$$ (This state can be represented by the array $$$[3, -1, -1]$$$)
3. Startup $$$2$$$ acquires startup $$$3$$$ (This state can be represented by the array $$$[-1, -1, 2]$$$).
4. Startup $$$2$$$ acquires startup $$$1$$$ (This state can be represented by the array $$$[2, -1, 2]$$$).

At this point, there is only one active startup, and this sequence of steps took $$$4$$$ days. It can be shown the expected number of days is $$$3$$$.

For the second sample, there is only one active startup, so we need zero days.

For the last sample, remember to take the answer modulo $$$10^9+7$$$.

Example input:
3
-1 -1 -1

Example output:
3

Example input:
2
2 -1

Example output:
0

Editorial:
You can solve this most likely by brute forcing small cases and looking for a pattern.
Here's a proof of the pattern. Let's define a potential of a startup with $$$k$$$ followers to be equal to $$$2^k - 1$$$. For example, an active startup with zero followers has potential zero.
Now, in one day, if we choose a startup with $$$p$$$ followers, and a startup with $$$q$$$ followers, the expected change in sum of potentials is equal to $$$\frac{1}{2}((2^{p+1} - 1 - 2^p + 1) - 2^q + 1) + \frac{1}{2}((2^{q+1} - 1 - 2^q + 1) - 2^p + 1) = 1$$$, regardless of the values of $$$p$$$ and $$$q$$$.
The potential of the final state is $$$2^{n-1} - 1$$$, and we can compute the current potential in linear time, so the expected number of turns is the difference between them.
To prove this more rigorously, we can use show that this process is a martingale, so we can use https://en.wikipedia.org/wiki/Optional_stopping_theorem to show that the expected number of days is exactly equal to this difference.