Problem:
Score : 600 points

Problem Statement

Given are N points (x_i, y_i) in a two-dimensional plane.

Find the minimum radius of a circle such that all the points are inside or on it.

Constraints

2 \leq N \leq 50

0 \leq x_i \leq 1000

0 \leq y_i \leq 1000

The given N points are all different.

The values in input are all integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print the minimum radius of a circle such that all the N points are inside or on it.

Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.

Sample Input 1

2
0 0
1 0

Sample Output 1

0.500000000000000000

Both points are contained in the circle centered at (0.5,0) with a radius of 0.5.

Sample Input 2

3
0 0
0 1
1 0

Sample Output 2

0.707106781186497524

Sample Input 3

10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9

Sample Output 3

6.726812023536805158

If the absolute or relative error from our answer is at most 10^{-6}, the output will be considered correct.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define endl '\n'
#define fast ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define zy -2147382637
#define bql 2147483647
#define ll long long
#define ull unsigned long long
#define ld long double
#define il inline
#define sz(x) x.size()
#define maxn 100010
#define rp(i, l, r) for (int i = l; i <= r; i++)
#define rb(i, r, l) for (int i = r; i >= l; i--)
#define fi first
#define se second
#define pb push_back
#define mp make_pair

using namespace std;
ll max(ll a,ll b){if(a>b)return a;else return b;}
ll min(ll a,ll b){if(a<b)return a;else return b;}
ll lowbit(ll x){return x&(-x);}
ll prime(ll x){
   if(x<=1)return false;
   for(int i=2;i<=int(sqrt(x));i++){
	   if(x%i==0)return false;
   }return true;
}
bool cmp(ll a,ll b){return a>b;}
ll gcd(ll a,ll b){ll r;while(b>0){r=a%b;a=b;b=r;}return a;}
ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
inline ll read() {
	char c = getchar(); ll w = 1; ll x = 0;
	while (!isdigit(c))
		(c == '-') && (w = -w), c = getchar();
	while (isdigit(c))
		x = (x << 1) + (x << 3) + (c ^ '0'), c = getchar();
	x *= w;
}
void fl(string name){
	freopen((name+".in").c_str(),"r",stdin);
	freopen((name+".out").c_str(),"w",stdout);
}
const int dx[8] = {-1, 1, 0, 0, -1, 1, -1, 1},
		  dy[8] = {0, 0, -1, 1, -1, -1, 1, 1};
int main(){
	fast;
	ll n;
	cin>>n;
	long double maxx=0.0,maxy=0.0;
	rp(i,1,n){
		long double x,y;
		cin>>x>>y;
		maxx=max(maxx,x),maxy=max(maxy,y);
	}
	cout<<fixed<<setprecision(100)<<sqrt(maxx*maxx+maxy*maxy)/2<<endl;
}/*『你强归你强，Siyuan比你强』*/