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: Accepted)

import java.util.*;
import java.io.*;

public class Main {	
	private static void solve(){
		//Implement solution here.
		double A = nd();
		double B = nd();
		double H = nd();
		double M = nd();
		double ang = 30D * H + 0.5D * M - 6D * M;
		double asin = A * Math.sin(Math.toRadians(ang));
		double acos = A * Math.cos(Math.toRadians(ang));
		
		double height = asin;
		double width = B - acos;
		
		System.out.println(Math.sqrt(height * height + width * width));

	}

	//Switch input source (stdin/file)
	private static BufferedReader br;
	public static void main(String[] args){
		br = new BufferedReader(new InputStreamReader(System.in));
		String debugDataPath = System.getenv("DD_PATH");        
		if(debugDataPath != null){
			try{
				br = new BufferedReader(new InputStreamReader(new FileInputStream(new File(debugDataPath))));
			}catch(Exception e){
				throw new RuntimeException(e);
			}
		}
		boolean hasNext = true;
		while(hasNext){
			solve();	
			try{
				br.mark(1);
				if(br.read() > 0){
					br.reset();			
					System.out.println("---");
				} else {
					hasNext = false;
				}
			}catch(Exception e){
				throw new RuntimeException(e);
			} 
		}
	}

	//Input read utility
	private static StringTokenizer tokenizer = null;
	public static String next() {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}
	//Get next single int
	private static int ni() {
		return Integer.parseInt(next());
	}
	//Get next single long
	private static long nl() {
		return Long.parseLong(next());
	}
	//Get next single double
	private static double nd() {
		return Double.parseDouble(next());
	}
	//Get next int array from one line
	private static int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}
	//Get next char array from one line
	private static char[] ns() {
		return next().toCharArray();
	}
	//Get next long array from one line
	private static long[] nal(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nl();
		return a;
	}
}