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)

// 課題4_2 AtCoder Beginner Contest 168 C - : (Colon)
 
#include<stdio.h>
#include<math.h>
 
int main(void){
    // 時針の長さlength_h cm,分針の長さlength_min cm,時刻hour:min
    int length_h, length_min, hour, min;
    // 結果の格納ans, 短針と時針の間の角度rad
    double ans, rad;
 
    // length_h, length_min, hour, minを受け取る
    scanf("%d %d %d %d", &length_h, &length_min, &hour, &min);

    // intからdoubleに変換
    length_h = (double)length_h;
    length_min = (double)length_min;
    min = (double)min;
    hour = (double)hour;
 
    // 円周率M_PI
    // (分針)ー(時針)cosなので角度の正負は気にしない
    rad = 2.0 * M_PI / 60.0 * min - 2.0 * M_PI / 12 * (hour + min / 60.0);
    // 余弦定理 ans^2
    ans = length_h * length_h + length_min * length_min - 2.0 * length_h * length_min * cos(rad);
    // 平方根をとる
    ans = sqrt(ans);

    printf("%.20f\n", ans);
    
    return 0;
}