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<(int)n;i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;

int n;
vector<int> X, Y;

double g(double x, double y) {
    double ma = 0;
    rep(i, n) {
        ma = max(ma, sqrt((X[i] - x) * (X[i] - x) + (Y[i] - y) * (Y[i] - y)));
    }
    return ma;
}

double f(double x) {
    double low_y = 0, high_y = 1000;
    while (high_y - low_y > 1e-8) {
        double c1 = (low_y * 2 + high_y) / 3, c2 = (high_y * 2 + low_y) / 3;
        if (g(x, c1) < g(x, c2)) high_y = c2;
        else low_y = c1;
    }
    return g(x, high_y);
}

int main() {
    cin >> n;
    rep(i, n) {
        int a, b;
        cin >> a >> b;
        X.push_back(a), Y.push_back(b);
    }
    double low_x = 0, high_x = 1000;
    while(high_x - low_x > 1e-8) {
        double c1 = (low_x * 2 + high_x) / 3, c2 = (high_x * 2 + low_x) / 3;
        if (f(c1) < f(c2)) high_x = c2;
        else low_x = c1;
    }
    cout << fixed << setprecision(8) << f(low_x) << endl;
    return 0;
}
