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, n) for (ll i = 0; i < ll(n); i++)
typedef pair<double, double> P;

int n;
vector<P> points;
double res;

double score(P p) {
    double ans = 0;
    rep(i, n) {
        double xdiff = p.first - points[i].first;
        double ydiff = p.second - points[i].second;
        double dist = sqrt(xdiff * xdiff + ydiff * ydiff);
        ans = max(ans, dist);
    }
    return res = ans;
}

double solvey(double x) {
    double l = 0, r = 1000;
    rep(i, 100) {
        double ly = (l * 2 + r) / 3;
        double ry = (l + r * 2) / 3;
        if (score(P(x, ly)) > score(P(x, ry))) {
            l = ly;
        } else {
            r = ry;
        }
    }
    return l;
}

int main() {
    cin >> n;
    rep(i, n) {
        int x, y;
        cin >> x >> y;
        points.emplace_back(x, y);
    }
    double l = 0, r = 1000;
    rep(i, 100) {
        double lx = (l * 2 + r) / 3;
        double rx = (l + r * 2) / 3;
        double ly = solvey(lx);
        double ry = solvey(rx);
        if (score(P(lx, ly)) > score(P(rx, ry))) {
            l = lx;
        } else {
            r = rx;
        }
    }
    printf("%.9f\n", res);
}