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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define sz(x) int(x.size())
#define show(x) {for(auto i: x){cout << i << " ";} cout << endl;}
using namespace std;
using ll = long long;
using P = pair<int, int>;

int N;
vector<int> X, Y;

double dist(double x, double y) {
    double ma = 0;
    for (int i = 0; i < N; i++) {
        double dx = X[i] - x;
        double dy = Y[i] - y;
        ma = max(ma, dx * dx + dy * dy);
    }
    return ma;
}

double g(double x) {
    double left = 0, right = 1000;
    rep(i, 80) {
        double l2 = (left * 2 + right) / 3;
        double r2 = (left + right * 2) / 3;
        if (dist(x, l2) < dist(x, r2)) right = r2;
        else left = l2;
    }
    return dist(x, left);
}

int main() {
    // 2重3分探索
    cin >> N;
    X = Y = vector<int> (N);
    rep(i, N) {
        cin >> X[i] >> Y[i];
    }
    double left = 0, right = 1000;
    while (abs(right - left) > 0.00000001) {
        double l2 = (left * 2 + right) / 3;
        double r2 = (left + right * 2) / 3;
        if (g(l2) < g(r2)) right = r2;
        else left = l2;
    }
    double ans = sqrt(g(left));
    printf("%.10f\n", ans);
    return 0;
}