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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <stdlib.h>
#include <string.h>
#include <math.h>
using namespace std;
#define ll long long                                    // long long省略
#define pb push_back                                    // push_back省略
#define mp make_pair                                    // make_pair省略
#define fi first                                        // first省略
#define se second                                       // second省略
#define itn int                                         // int誤字保険
#define count cout                                      // cout誤字保険
#define vecotr vector                                   // vector誤字保険
#define opt() cin.tie(0); ios::sync_with_stdio(false)   // 入出力速度改善

#define rep(i,l,r)for(ll i=(l);i<(r);i++)               // 範囲[l, r)で刻み1のfor文(順方向)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))         // 範囲[l, r)で刻みkのfor文(順方向)
#define rrep(i,l,r)for(ll i=(r-1);i>=(l);i--)           // 範囲[l, r)で刻み1のfor文(逆方向)
#define rrepp(i,l,r,k)for(ll i=(r-1);i>=(l);i-=(k))     // 範囲[l, r)で刻みkのfor文(逆方向)
#define all(x) (x).begin(), (x).end()                   // vectorのポインタ位置指定用
#define max(p,q)((p)>(q)?(p):(q))                       // max拡張
#define min(p,q)((p)<(q)?(p):(q))                       // min拡張
#define bit(n,m)(((n)>>(m))&1)                          // 変数nのm番目のbitを取り出す

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 (b<a) { a=b; return 1; } return 0; }

int dy[]={0, 0, 1, -1};                                 // 4方向近傍
int dx[]={1, -1, 0, 0};                                 // 4方向近傍


/*
   upll : (a<b)なら-1, (a=b)なら0, (a>b)なら1
 downll : (a<b)なら1, (a=b)なら0, (a>b)なら-1

 const void* a : 型を縛らずにconstをつける(=変数aが読み込み専用の変数であることを明示する)
        (ll*)a : 変数aがconst long long *型であると明示する
*/
// int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
// int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
// void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
// void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
ll pom(ll a,ll n,int m){ll x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}

#define INF ((1LL<<62)-(1LL<<31))
//#define MOD 998244353
#define MOD 1000000007
#define invp(a,p) pom(a,p-2,p)

int main(){
    const double PI=3.14159265358979323846;
    double a, b, x;
    cin >> a >> b >> x;
    double n = 2*x / (b * a);
    if(n >= a){
        double n1 = (2*b - 2*x/(a*a));
        cout << 90 - (180/PI) * atan(a/n1) << endl;
    }else{
        cout << 180*atan(b/n) / PI << endl;
    }
    
}