Particle Collision

Building a particle collider is hard, and running one experiment on it costs a lot. Checking an idea in simulation first is cheaper than testing it on real hardware. Write a simple simulator for the setup below.

The system holds exactly three particles. All of them stay on an infinite plane, so each one is a circle. The position of particle $i$ is given only by the coordinates $x_i$ and $y_i$ of its center ($1 \le i \le 3$). The three particles share the same radius $r$, and all of them are at rest at the start.

Once the experiment starts, particle 1 moves in the direction of the vector $(x_v, y_v)$. When particle $i$ touches particle $j$, particle $j$ starts moving in the direction perpendicular to the common tangent at the contact point, which points away from particle $i$. At that instant particle $i$ disappears and turns into radiation. A moving particle that touches no other particle keeps moving forever.

The simulator has to tell these five outcomes apart.

1. Particle 1 touches particle 2, and particle 2 then touches particle 3.
2. Particle 1 touches particle 3, and particle 3 then touches particle 2.
3. Particle 1 touches particle 2, and particle 2 moves forever.
4. Particle 1 touches particle 3, and particle 3 moves forever.
5. Particle 1 moves forever.

The input has four lines. Each of the first three lines holds two integers $x_i$ and $y_i$ ($|x_i|, |y_i| \le 1000$), describing particle 1, particle 2, and particle 3 in that order. The fourth line holds three integers $x_v$, $y_v$, and $r$ ($|x_v|, |y_v| \le 1000$, $0 < r \le 50$). The vector $(x_v, y_v)$ is not the zero vector.

No two particles touch or overlap at the start, and the distance between the centers of particle 2 and particle 3 is greater than $4r$.

Print the number of the outcome listed above as a single integer from 1 to 5.

Compute carefully. The outcome is guaranteed to stay the same if the initial vector $(x_v, y_v)$ is rotated by one degree in either direction.