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>
#include <fstream>

using namespace std;
#define intMax 2147483647
#define rep(i, a, b) for (int i = a; i < b; i++)
#define repp(i, a, b) for (int i = a; i >= b; i--)
#define be(a) (a.begin(), a.end())
#define rbe(p) (p.rbegin(), p.rend())
#define pb push_back

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int n;
    cin >> n;
    vector<pair<int, int>> A;
    rep(i, 0, n)
    {
        int a, b;
        cin >> a >> b;
        A.pb({a, b});
    }
    long double maxi = 0;
    int l = 0, r = 0;
    rep(i, 0, n)
    {
        rep(j, 0, n)
        {
            if (l + r < abs(A[i].first - A[j].first) + abs(A[j].second - A[i].second))
            {
                l = abs(A[i].first - A[j].first);
                r = abs(A[j].second - A[i].second);
            }
        }
    }
    maxi = (double)(sqrt((l * l) + (r * r)));
    cout << fixed << setprecision(20) << (double)(maxi/2.0);
}
/*
20 20
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*/