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 C (GCC 9.2.1) (judge verdict: Accepted)

/* ex4_2 KitamoriFumiya */
#include <stdio.h>
#include <math.h>
#define PI 3.14159265359
 
//  余弦定理で長針と短針の先端の距離を出力する関数
void cosine(double length_hour, double length_minute, double hour, double minute){
	double angle_hour, angle_minute, difference_angle; // angle_hour(長針の角度), angle_minute(短針の角度), difference_angle(長針と短針の角度の差)をdouble型の変数として宣言
	angle_hour = 360 * hour / 12 + 30 * minute / 60; // 長針の角度を計算
	angle_minute = 360 * minute / 60; // 短針の角度を計算
	difference_angle = angle_hour - angle_minute; // 長針と短針の角度の差を計算
	difference_angle = (difference_angle / 180) * PI ; // 角度の差を弧度法で表す
	double distance; // 長針と短針の先端の距離をdouble型の変数として宣言
	distance = length_hour * length_hour + length_minute * length_minute - 2 * length_hour * length_minute * cos(difference_angle); // 距離の2乗を余弦定理で求める
	distance = sqrt(distance); // 距離をルートに入れて戻す
	printf("%.10lf\n", distance); // 距離を少数第10位まで出力する
}
 
int main(void){
	double length_hour, length_minute, hour, minute; // length_hour(長針の長さ), length_minute(短針の長さ), hour(時間), minute(分数)をdouble型の変数として宣言
	scanf("%lf %lf %lf %lf", &length_hour, &length_minute, &hour, &minute); // length_hour, length_minute, hour, minuteを入力
	
	cosine(length_hour, length_minute, hour, minute); // 余弦定理で長針と短針の先端の距離を出力する関数を実行
	
	return 0;
}