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: Accepted)

#define _USE_MATH_DEFINES
#include<iostream>
#include<string>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<list>
#include<iomanip>
#include<vector>
#include<random>
#include<functional>
#include<algorithm>
#include<stack>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<unordered_map>
#include<climits>
#include<fstream>
#include<time.h>
#include<cassert>
#include<functional>
#include<numeric>
#include <tuple>

using namespace std;
using ll = long long;
using ld = long double;
#define int long long
#define all(a) (a).begin(),(a).end()
#define EPS (1e-7)
#define fs first
#define sc second
#define xx first
#define yy second.first
#define zz second.second
#define H pair<ld, ld>
#define P pair<int, pair<int, int>>
#define Q(i,j,k) mkp(i,mkp(j,k))
#define rep(i,n) for(int (i) = 0 ; (i) < (n) ; (i)++)
#define rng(i,s,n) for(int (i) = (s) ; (i) < (n) ; (i)++)
#define req(i,n) for(int (i) = (n) - 1 ; (i) >= 0 ; (i)--)
#define mkp make_pair
#define mem0(x) memset(x,0,sizeof(x))
#define vec vector
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define crdcomp(b) sort(all((b)));(b).erase(unique(all((b))),(b).end())
#define ssp(i,n) (i==n-1?"\n":" ")
#define ctoi(c) (int)(c-'A')
#define itoc(c) (char)(c+'A')
#define pb push_back
#define cauto const auto&
#define itn int
#define pri_que priority_queue
#define pp(x,y) pb(H{x,y})
#define ppp(x,y,z) pb(Q(x,y,z))
const int mod = 1e9 + 7;
const int Mod = 998244353;
const ll inf = 3 * 1e18;
const int Inf = 15 * 1e8;
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; }
ll read() { ll u, k = scanf("%lld", &u); return u; }
string reads() { string s; cin >> s; return s; }
ll gcd(ll i, ll j) { return j ? gcd(j, i % j) : i; }
ll mod_pow(ll x, ll n, ll p) {
	ll res = 1;
	while (n > 0) {
		if (n & 1) res = res * x % p;
		x = x * x % p;
		n >>= 1;
	}
	return res;
}//x^n%p
ll bitcount(ll x) {
	int sum = 0; for (int i = 0; i < 60; i++)if ((1ll << i) & x) sum++;
	return sum;
}
const int fn_ = 600000;
ll fact_[fn_], comp_[fn_];
ll comb(ll x, ll y, ll Mod = mod) {
	if (!fact_[0]) {
		fact_[0] = 1; comp_[0] = 1;
		for (int i = 1; i < fn_; i++) {
			fact_[i] = fact_[i - 1] * i % Mod;
			comp_[i] = mod_pow(fact_[i], Mod - 2, Mod);
		}
	}
	if (x < y) return 0;
	return fact_[x] * comp_[x - y] % Mod * comp_[y] % Mod;
}
//---------------------------------------------------------------
int n;
vec<H>a;
ld dist(H a, H b) {
	return sqrt((a.fs - b.fs) * (a.fs - b.fs) + (a.sc - b.sc) * (a.sc - b.sc));
}
signed main() {
	cin >> n;
	rep(i, n) {
		a.pp(read(), read());
	}
	ld ans = Inf;
	rep(i, n) {
		rng(j, i + 1, n) {
			rng(k, j + 1, n) {
				H t;
				t.sc = (a[k].fs - a[i].fs) * (a[i].fs * a[i].fs + a[i].sc * a[i].sc - a[j].fs * a[j].fs - a[j].sc * a[j].sc) -
					(a[j].fs - a[i].fs) * (a[i].fs * a[i].fs + a[i].sc * a[i].sc - a[k].fs * a[k].fs - a[k].sc * a[k].sc);
				t.sc /= 2 * (a[k].fs - a[i].fs) * (a[i].sc - a[j].sc) - 2 * (a[j].fs - a[i].fs) * (a[i].sc - a[k].sc);
				if (a[i].fs == a[j].fs) {
					t.fs = 2 * (a[i].sc - a[k].sc) * t.sc - a[i].fs * a[i].fs - a[i].sc * a[i].sc + a[k].fs * a[k].fs + a[k].sc * a[k].sc;
					t.fs /= 2 * (a[k].fs - a[i].fs);
				}
				else {
					swap(j, k);
					t.fs = 2 * (a[i].sc - a[k].sc) * t.sc - a[i].fs * a[i].fs - a[i].sc * a[i].sc + a[k].fs * a[k].fs + a[k].sc * a[k].sc;
					t.fs /= 2 * (a[k].fs - a[i].fs);
					swap(j, k);
				}
				ld r = dist(t, a[i]);
				rep(g, n) {
					if (dist(t, a[g]) > r + EPS) {
						goto loop;
					}
				}
				ans = min(ans, r);
			loop:;
			}
		}
	}
	rep(i, n) {
		rng(j, i + 1, n) {
			H t = H{ a[i].fs + a[j].fs,a[i].sc + a[j].sc };
			t.fs /= 2;
			t.sc /= 2;
			ld r = dist(t, a[i]);
			rep(g, n) {
				if (dist(t, a[g]) > r + EPS) {
					goto roop;
				}
			}
			ans = min(ans, r);
		roop:;
		}
	}
	printf("%.20LF\n", ans);
}