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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
#define cinf(n,x) for(int i=0;i<(n);i++)cin>>x[i];
#define ft first
#define sc second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(v) (v).begin(),(v).end()
#define LB(a,x) lb(all(a),x)-a.begin()
#define UB(a,x) ub(all(a),x)-a.begin()
#define mod 1000000007
#define FS fixed<<setprecision(15)
using namespace std;
typedef long long ll;
template<class T> using V=vector<T>;
using Graph = vector<vector<int>>;
using P=pair<ll,ll>;
typedef unsigned long long ull;
typedef long double ldouble;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

const ll INF=1e18;
//abc168

int main(){
    cin.tie(0);ios::sync_with_stdio(false);
    ll a,b,h,m;
    cin>>a>>b>>h>>m;
    double s=(double)360*h/12+30*m/60;
    double t=360*m/60;
    double u=abs(s-t);
    if(u>180) u=360-u;
    cout<<FS<<(double)sqrt(a*a+b*b-2*a*b*cos(u*M_PI/180))<<endl;
}
//ペナルティ出しても焦らない　ACできると信じろ！！！
//どうしてもわからないときはサンプルで実験　何か見えてくるかも
//頭で考えてダメなら紙におこせ！！
/*
V,P(大文字)使用不可
乗算などの際にオーバーフローに注意せよ！
(適切にmodをとれ　にぶたんで途中で切り上げろ)
制約をよく読め！
{
    ・全探索できるなら全探索しろ
    ・異常な制約ならそこに注目
}
stringの計算量(扱い)注意
コーナー注意！(特に数値が小さいものについては要検証)
N行出力のときは'¥n'
グリッド上では行先が範囲内におさまるかif文で確認(RE注意)
if文ではちゃんと比較演算子==を使え('='でもエラー出ない)
配列(vector)の大きさが0か1以上かで場合分けせよ(RE注意)
(vector<int> a(m)でm==0というものはできない)
modはなるべく最後に取れ！
*/