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

int main() {
    uint N;
    cin >> N;
    vector<double> X(N);
    vector<double> Y(N);
    double lx = 1000;
    double hx = 0;
    double ly = 1000;
    double hy = 0;

    for (uint i = 0; i < N; i++) {
        cin >> X[i] >> Y[i];
        lx = min(lx, X[i]);
        hx = max(hx, X[i]);
        ly = min(ly, Y[i]);
        hy = max(hy, Y[i]);
    }

    double curX = (hx + lx) / 2;
    double curY = (hy + ly) / 2;
    double nxtX = 0;
    double nxtY = 0;

    double alfa = 0.1;
    uint idx = 0;

    while (alfa > pow(10, -8)) {
        curX = nxtX;
        curY = nxtY;
        double maxD = 0;

        for (uint i = 0; i < N; i++) {
            double d = pow(X[i] - curX, 2) + pow(Y[i] - curY, 2);

            if (maxD < d) {
                maxD = d;
                idx = i;
            }
        }

        nxtX = (1 - alfa) * curX + alfa * X[idx];
        nxtY = (1 - alfa) * curY + alfa * Y[idx];
        alfa = alfa * 0.9999;
    }

    double ans = sqrt(pow(X[idx] - curX, 2) + pow(Y[idx] - curY, 2));
    printf("%.6f\n", ans);
}
