Problem:
Score: 300 points

Problem Statement

Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.

An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.

At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?

Constraints

All values in input are integers.

1 \leq A, B \leq 1000

0 \leq H \leq 11

0 \leq M \leq 59

Input

Input is given from Standard Input in the following format:

A B H M

Output

Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.

Sample Input 1

3 4 9 0

Sample Output 1

5.00000000000000000000

The two hands will be in the positions shown in the figure below, so the answer is 5 centimeters.

Sample Input 2

3 4 10 40

Sample Output 2

4.56425719433005567605

The two hands will be in the positions shown in the figure below. Note that each hand always rotates at constant angular velocity.

Solution in Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
    // １分間に長針が動く角度
    private static double LONG_ANGLE  = 0.5;
    // １分間に短針が動く角度
    private static double SHORT_ANGLE = 6.0;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        // 長針の長さ
        int a = sc.nextInt();
        // 短針の長さ
        int b = sc.nextInt();
        
        int h = sc.nextInt();
        int m = sc.nextInt();
        
        System.out.println(
                Math.sqrt(a * a + b * b - 2 * a * b * Math.cos(Math.toRadians(shortAngle(h, m)))));
    }

    /**
     * 指定した時間の長針と短針が作り出す小さい角度を返します。
     *
     * @param h 時間
     * @param m 分
     * @return 短針と長針が作り出す角度
     */
    private static double shortAngle(int h, int m) {
        double a = LONG_ANGLE * (h * 60 + m);
        double b = SHORT_ANGLE * m;

        if (Math.abs(a - b) >= 180) {
            return Math.abs(a - b) - 180.0;
        }

        return Math.abs(a - b);
    }
}