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;

#define rep(i, n) for(int i = 0; i < (n); ++i)

typedef long long ll;
typedef pair<ll, ll> P;

const int MAX = 1e5 + 5;
const int INF = 1001001001;
const int MOD = 1e9 + 7;

int N;
P dot[55];

double dist(double x, double y) {
    double res = 0;
    rep(i, N) {
        res = max(res, (dot[i].first - x) * (dot[i].first - x) + (dot[i].second - y) * (dot[i].second - y));
    }
    return sqrt(res);
}

double f(double x) {
    double l = 0, r = INF;
    rep(i, 100) {
        double y1 = (l * 2 + r) / 3;
        double y2 = (l + r * 2) / 3;
        if (dist(x, y1) < dist(x, y2)) {
            r = y2;
        } else {
            l = y1;
        }
    }
    return dist(x, l);
}

int main(){
    cin >> N;

    rep(i, N) {
        int x, y;
        cin >> x >> y;
        dot[i] = P(x, y);
    }

    double l = 0, r = INF;
    rep(i, 100) {
        double x1 = (l * 2 + r) / 3;
        double x2 = (l + r * 2) / 3;
        if (f(x1) < f(x2)) {
            r = x2;
        } else {
            l = x1;
        }
    }

    printf("%.10f\n", f(l));
}