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: Wrong Answer)

/* ex4_2
    fmuacshai */

#include<stdio.h>
#include<math.h>
int main (void){
    //時針の長さをh_hand、分針の長さをm_hand、○時をhour、×分をminと定義する
    double h_hand,m_hand,hour,min;
    //時針と分針の角度をangle、求める長さの2乗をlenth2、求める長さをans
    double angle,ans,lenth2;
    //h_handを入力
    scanf("%lf",&h_hand);
    //m_handを入力
    scanf("%lf",&m_hand);
    //hourを入力
    scanf("%lf",&hour);
    //minを入力
    scanf("%lf",&min);
    //angleを計算（ラジアン）
    angle=fabs(30*hour-11*min/2)/180*M_PI;
    //lenth2を予言定理で求める
    lenth2=pow(h_hand,2)+pow(m_hand,2)-2*h_hand*m_hand*cos(angle);
    //ansを求める
    ans=sqrt(lenth2);
    //ansを出力
    printf("%f\n",ans);
    return 0;
}