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)

/***********************************************************
 *This code By @1353055672(Ligen)                          *
 *[Warning]You're not excepted to understand this code!    *
 *CSP-S 2019 RP++                                          *
 ***********************************************************/
// #pragma GCC optimize(3)
#define ADD_STACK int size = 512 << 20;\
	char *pp = (char*)malloc(size) + size;  \
	__asm__("movl %0, %%esp\n" :: "r"(pp))
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<fstream>
#include<queue>
#include<map>
#include<stack>
#include<set>
#include<bitset>
#include<ctime>
#include<vector>
#define ll long long
#define ull unsigned long long
#define mn 200020
#define Max(x,y) (((x)>(y))?(x):(y))
#define Min(x,y) (((x)<(y))?(x):(y))
#define Abs(x) (((x)<(0))?(-(x)):(x))
#define infll (ll)(1e18)
#define infint (1<<30)
#define mod (int)(1e9+7)
#define FOR(a,b,c) for (register int a=b;a<=c;++a)
#define FORD(a,b,c) for (register int a=b;a>=c;--a)
using namespace std;
// char buf[1<<20],*p1=buf,*p2=buf;
// #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?EOF:*p1++)
inline ll read(){
	ll x=0,f=1;char c;
	for(c=getchar();c<'0'||c>'9';f=((c=='-')?-1:f),c=getchar());
	for(;c>='0'&&c<='9';x=x*10+c-'0',c=getchar());
	return x*f;
}
template<typename T>
inline void write(T a){
	if(a==0){putchar('0');return;}if(a<0)putchar('-'),a=-a;char c1[120];int h=0;
	while(a)c1[++h]=a%10+'0',a/=10;FORD(i,h,1)putchar(c1[i]);
}
inline void write_(){return;}
template<typename T,typename... Args>
inline void write_(T a,Args... b){write(a);putchar(' ');write_(b...);}
inline void writeln(){putchar('\n');return;}
template<typename T,typename... Args>
inline void writeln(T a,Args... b){write(a);putchar(' ');writeln(b...);}
//need c++11

//inline void write(ll a){
//    if(a==0){putchar('0');return;}if(a<0)putchar('-'),a=-a;char c1[120];int h=0;
//    while(a)c1[++h]=a%10+'0',a/=10;FORD(i,h,1)putchar(c1[i]);
//}
//inline void write_(ll a){write(a);putchar(' ');}
//inline void writeln(ll a){write(a);putchar('\n');}
inline ll gcd(ll a,ll b){return a==0?b:gcd(b%a,a);}
inline ll lcm(ll a,ll b){return 1ll*a/gcd(a,b)*b;}
inline ll Pow(ll n,ll a){ll b=1;while(a){if(a&1)b=1ll*b*n%mod;n=1ll*n*n%mod;a>>=1;}return b;}
//---------------------Head Files--------------------------//
#define pi 3.14159265358979323
signed main(){
#ifdef LOCAL_LIGEN
	freopen("0.in","r",stdin);
	// freopen("0.out","w",stdout);
	double be=clock();
#endif
	int a=read(),b=read(),x=read();
	if(a*a*b==x)return puts("90.0000000000000000000"),0;
	double xx=x/(1.0000000000000000000000000*a);
	if(xx>=(a*b/2.00000000000000000))printf("%.10lf\n",90.0000000000000000000-((180.00000000000000000000000000/pi)*atan2(a*a,(2.000000000000000*(a*b-xx)))));else
	printf("%.10lf\n",(180.00000000000000000000000000/pi)*atan2(b*b,(2.000000000000000*(xx))));
#ifdef LOCAL_LIGEN
	double en=clock();
	printf("Time: %.0lfms\n",en-be);
	fclose(stdin);fclose(stdout);
#endif
	return 0;
}