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 <bits/stdc++.h>
#define rep(i,n)    for(long long i = 0; i < (n) ; ++i)
#define orep(i,n)   for(long long i = 1; i <= (n); ++i)   // one rep
#define drep(i,n)   for(long long i = (n)-1; i >= 0; --i) // down rep
#define srep(i,s,t) for(long long i = (s); i < (t); ++i)  // set rep
#define rng(x)  (x).begin(),(x).end()

  // range
#define rrng(a) (a).rbegin(),(a).rend() // reverse range
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define ru(x,y) (((x)+(y)-1)/(y)) // round up
#define fi first
#define se second
#define eb emplace_back
#define fcout cout << fixed << setprecision(15)


using namespace std;
using ll  = long long;
using P   = pair<int,int>;
using vi  = vector<int>;
using vvi = vector<vi>;
using vp  = vector<P>;
using vl  = vector<ll>;

template<typename T> void Yes(T flag) {
  cout << (flag ? "Yes" : "No") << endl;
}

template<typename T> void pv(vector<T> vec) {
  cout << "[";
  for(auto& v : vec) { cout << v << ","; }
  cout << "]" << endl;
}

template<>
void pv(vector<P> vec) {
  cout << "[";
  for(auto& v : vec) {
    cout << "<" << v.fi << ","<< v.se << ">" << ",";
  }
  cout << "]" << endl;
}

template<typename T> void pv2(vector<vector<T>> vec) {
  for(auto& v : vec) pv(v);
}

static const double pi = 3.141592653589793;

int main() {
  int zi,hun;
  cin >> zi >> hun;
  int h,m;
  cin >> h >> m;
  double theta=abs(m/60.0-(h/12.0+m/720.0))*2*pi;
  fcout << sqrt(zi*zi+hun*hun-2*zi*hun*cos(theta)) << endl;


  return 0;
}