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 melo */
#include <stdio.h>
#include <math.h>//数学関数のヘッダーファイルを読み込む
#define PI 3.141592653589793238//円周率PIを十分な長さ定義する
int main(void){
    int lenN,lenM,hour,min;//整数の箱を四つ用意する
    double angleN,angleM,angleN_M,lenN_M;//倍精度の実数の箱を四つ用意する
    scanf("%d%d%d%d",&lenN,&lenM,&hour,&min);//時針の長さ、分針の長さ、H時、M分を代入する
    angleN=(2*PI/12)*hour+(2*PI/720)*min;//時針の0時から進んだ角度
    angleM=(2*PI/60)*min;//分針の0分から進んだ角度
    angleN_M=angleN-angleM;//時針と分針のなす角度（余弦なので-をとってもPIを超えていてもx軸対象なので問題ない）
    lenN_M=sqrt(lenN*lenN+lenM*lenM-2*lenN*lenM*cos(angleN_M));//余弦定理から、三角形の各辺A,B,Cと角ｃでC^2=A^2+B^2-2ABcos(c)が成り立つことを利用する
    printf("%.18lf\n",lenN_M);
    return 0;
}