Runner and Sniper

You are running from an enemy. The enemy is a sniper with a laser gun, and a single hit ends the chase, so you want to know how fast you have to run to stay alive.

You and the sniper are on the xy-plane, with the x-axis pointing right and the y-axis pointing up. The plane is infinite, and nothing blocks the laser or your movement.

The sniper and the gun are at $(0, 0)$ and never move. The gun turns continuously at a rate of at most $\omega$ degrees per unit time, clockwise or counterclockwise, and the sniper may reverse the turn at any moment. At time $0$ the gun points $\theta$ degrees counterclockwise from the positive direction of the x-axis.

You start at $(x, y)$ and may move in any direction at a speed of at most $v$. You choose the value of $v$ yourself. You are shot exactly when the gun points at your position, so the time the laser needs to reach you is ignored. You are a point, and the laser is a half-line whose endpoint is $(0, 0)$.

Both you and the sniper act optimally. Find the largest speed $v$ for which the sniper still shoots you in finite time.

One line holds four integers $x$, $y$, $\theta$, $\omega$ separated by spaces.

• $|x| \le 1000$, $|y| \le 1000$, and $(x, y) \ne (0, 0)$ give your starting position.
• $0 \le \theta < 360$ is the initial direction of the gun, measured counterclockwise from the positive x-axis in degrees.
• $1 \le \omega \le 100$ is the angle in degrees that the gun can turn per unit time.

At time $0$ the gun does not point at your position.

Print the largest speed $v$ on one line, rounded to exactly 8 digits after the decimal point, in the format printf("%.8f") produces. No test is close enough to a rounding boundary for a correct double-precision computation to change the last digit.