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## Learning Opportunities

This puzzle can be solved using the following concepts. Practice using these concepts and improve your skills.

## Goal

Spaceship is orbiting the planet in the Discrete Chebyshev 2D Space. By given initial spaceship position and velocity predict its position after time seconds.

Physics

Planet has its center in (0, 0). All points with Chebyshev distance to planet center less than or equal to radius, belongs to planet. Chebyshev distance between (x1, y1) and (x2, y2) is Max(Abs(x1 - x2), Abs(y1 - y2))

Spaceship works as follows:

1. Every turn spaceship moves if it is not crashed yet: velocity vector is added to ship position.
2. Spaceship crashed if its final position has chebyshev distance to planet center less than or equal to radius.
3. Chebyshev gravity is applied. Denote ship position as (x, y) and velocity as (vx vy). Then vx (and vy) is decreased by 1 if x > 0 (respectively y > 0) and increased by 1 if x < 0 (respectively y < 0).

Sample
Spaceship coordinates sequence after simulation for planet radius 1 and start velocity (2, 0):

(0; −3) → (2; −3) → (3; −2) → (3; 0) → (2; 2) → (0; 3) → (−2; 3) → (−3; 2) → (−3; 0) → (−2; −2) → ...

See image with this trajectory below.
`.............0.1....9....2....PPP....8.PPP.3....PPP....7....4....6.5.............`

P — is planet point, number — ship position after that number of seconds. Center of the planet is at (0; 0)
Input
Single line with 6 space separated integers: radius, x, y, vx, vy, time
Output
Single line with 3 space separated integers: x y crashed
Crashed is 1 if spaceship is crashed in the planet and 0 otherwise.
Constraints
-50 ≤ x, y, vx, vy ≤ 50
`1 0 -3 2 0 9`
`-2 -2 0`