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)

#include<cstdio>
#include<iostream>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<deque>
#include<tuple>
#include<algorithm>
#include<numeric>
#include<limits>
#include<boost/multiprecision/cpp_int.hpp>
using namespace boost::multiprecision;
using namespace std;
#define BEGIN ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define END return EXIT_SUCCESS
#define up(I,A,B) for(auto I=(A);I<=(B);++I)
#define dw(I,A,B) for(auto I=(A);I>=(B);--I)
#define rep(I,N) up(I,0,N-1)
#define all(C) (C).begin(),(C).end()
#define rall(C) (C).rbegin(),(C).rend()
#define ft first
#define sd second
#define mp make_pair
#define mt make_tuple
#define pf push_front
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define rs resize
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template<class T>inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T>inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T>inline void                                      in(T &p){cin>>p;}
template<class T1,class T2>inline void                            in(T1 &p1,T2 &p2){cin>>p1>>p2;}
template<class T1,class T2,class T3>inline void                   in(T1 &p1,T2 &p2,T3 &p3){cin>>p1>>p2>>p3;}
template<class T1,class T2,class T3,class T4>inline void          in(T1 &p1,T2 &p2,T3 &p3,T4 &p4){cin>>p1>>p2>>p3>>p4;}
template<class T1,class T2,class T3,class T4,class T5>inline void in(T1 &p1,T2 &p2,T3 &p3,T4 &p4,T5 &p5){cin>>p1>>p2>>p3>>p4>>p5;}
template<class T>inline void                                      ins(T &p){for_each(all(p),in<decltype(p[0])>);}
template<class T>inline void                             out(T p){cout<<p<<"\n";}
template<class T1,class T2>inline void                   out(T1 p1,T2 p2){cout<<p1<<" "<<p2<<"\n";}
template<class T1,class T2,class T3>inline void          out(T1 p1,T2 p2,T3 p3){cout<<p1<<" "<<p2<<" "<<p3<<"\n";}
template<class T1,class T2,class T3,class T4>inline void out(T1 p1,T2 p2,T3 p3,T4 p4){cout<<p1<<" "<<p2<<" "<<p3<<" "<<p4<<"\n";}
inline void                                              out(ld p){printf("%.12Lf\n",p);}
template<class T>inline void outml(T p){for_each(all(p),out<decltype(p[0])>);}//ml=multi line
template<class T>inline void outol(T V){auto n=V.size();rep(i,n)cout<<V[i]<<" \n"[i==n-1];}//ol=one line
ld A,B,H,M;
inline void solve(void){
  in(A,B,H,M);
  out(sqrt(A*A+B*B-2*A*B*cos(2.0*((60*H+M)/720.0-M/60.0)*M_PI)));
}
int main(void){
  BEGIN;
  solve();
  END;
}