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}.

Examples

Input

2
0 0
1 0


Output

0.500000000000000000


Input

3
0 0
0 1
1 0


Output

0.707106781186497524


Input

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


Output

6.726812023536805158
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, srt, end) for (long long i = (srt); i < (long long)(end); i++)
#define ITR 100

ll N;
vector<ll> vx, vy;

double f(double x, double y){
    double ret = 0;
    rep(i, 0, N){
        double dx = x - vx[i];
        double dy = y - vy[i];
        ret = max(ret, sqrt(dx*dx + dy*dy));   
    }
    return ret;
}

double g(double x){
    double L = -1000, R = 1000;
    rep(itr, 0, ITR){
        double m1 = (L * 2 + R) / 3;
        double m2 = (L + R * 2) / 3;
        if(f(x, m1) > f(x, m2)) L = m1;
        else R = m2;
    }
    return f(x, L);
}

int main(){
    cin >> N;
    vx.resize(N);
    vy.resize(N);
    rep(i, 0, N) cin >> vx[i] >> vy[i];

    double L = -1000, R = 1000;
    for(ll itr = 0; itr < ITR; itr++){
        double m1 = (L * 2 + R) / 3;
        double m2 = (L + R * 2) / 3;
        if(g(m1) > g(m2)) L = m1;
        else R = m2;
    }

    printf("%.20lf\n", g(L));
    return 0;
}