Problem:
Score : 400 points

Problem Statement

Takahashi has a water bottle with the shape of a rectangular prism whose base is a square of side a~\mathrm{cm} and whose height is b~\mathrm{cm}. (The thickness of the bottle can be ignored.)

We will pour x~\mathrm{cm}^3 of water into the bottle, and gradually tilt the bottle around one of the sides of the base.

When will the water be spilled? More formally, find the maximum angle in which we can tilt the bottle without spilling any water.

Constraints

All values in input are integers.

1 \leq a \leq 100

1 \leq b \leq 100

1 \leq x \leq a^2b

Input

Input is given from Standard Input in the following format:

a b x

Output

Print the maximum angle in which we can tilt the bottle without spilling any water, in degrees.
Your output will be judged as correct when the absolute or relative error from the judge's output is at most 10^{-6}.

Sample Input 1

2 2 4

Sample Output 1

45.0000000000

This bottle has a cubic shape, and it is half-full. The water gets spilled when we tilt the bottle more than 45 degrees.

Sample Input 2

12 21 10

Sample Output 2

89.7834636934

This bottle is almost empty. When the water gets spilled, the bottle is nearly horizontal.

Sample Input 3

3 1 8

Sample Output 3

4.2363947991

This bottle is almost full. When the water gets spilled, the bottle is still nearly vertical.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

// includes
#include "bits/stdc++.h"
using namespace std;
// macros
#define ll long long
#define MOD 998244353 //100000000  //1000000007 //
#define pii pair<ll, ll>
#define piii pair<ll, pii>
#define sz(x) ((ll)(x).size())
#define ft first
#define sd second
#define pb push_back
#define rep(i, n) for(ll i = 0; i < n; i++)
#define repr(i, n) for(ll i = n-1; i >= 0; i--)
#define itr(it, x) for(auto it = x.begin(); it != x.end(); it++)
#define mem(a, b) memset(a, (ll)b, sizeof(a))
#define all(v) v.begin(), v.end()
#define allr(v) v.rbegin(), v.rend()
#define edge(v, x, y) v[x].pb(y); v[y].pb(x);
#define popc __builtin_popcount
#define ANS(s) {cout << s << "\n"; return;}
// functions
template <typename T> void unique(T& c){c.erase(std::unique(c.begin(), c.end()), c.end());}
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << "(" << p.first << ", " << p.second << ")"; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << "(" << itr->first << ", " << itr->second << ")"; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << "(" << itr->first << ", " << itr->second << ")"; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
// constants
const ll N = 1e5;
const ll M = 1e6;
const ll A = 1e7;
const ll inf = 1e9;
const long long linf = 1LL << 60;
const double er = 1e-10;
const double pi = 3.141592653589793238463;
const ll lx[4] = {0, 0, 1, -1};
const ll ly[4] = {1, -1, 0, 0};
const ll dx[8] = {0, 0, 1, -1, 1, -1, 1, -1};
const ll dy[8] = {1, -1, 0, 0, 1, 1, -1, -1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;

double angle(double a, double b){
    return (atan(a/b)*180.0 )/pi;
}
void solve()
{
    ll a, b, x;
    cin >> a >> b >> x;
    double t= a*a*b, area, h, w;
    if(2*x < t){
        area = x*2;
        h = b; w = sqrt(area/h); cout << angle(h, w) << endl;
    }
    else {
        area = t-x;area *= 2;
        h = a; w = (area)/ (h*h); cout << angle(w, h);
    }
    //

}
int main(int argc, char const* argv[])
{
    ll t = 1; //cin >> t;
    while(t--) {
        solve();
    }
    return 0;
}