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 <iostream>
#include <cstdio>
#include <vector>
#include <cmath>

using namespace std;

typedef long long ll;
typedef long double ld;

const ll M = 900000;
const ld X = 0.6;
const ld Y = 0.9999;
ll n;
ll x[55], y[55];
ld cx, cy;
ld now = X;

void f(ld p)
{
    ld mx = 0;
    ll mi = 0;
    for(ll i = 0; i < n; i++)
    {
        ld r = (cx - x[i]) * (cx - x[i]) + (cy - y[i]) * (cy - y[i]);
        if(mx < r) mx = r, mi = i;
    }
    cx += (x[mi] - cx) * p;
    cy += (y[mi] - cy) * p;
}

int main()
{
    scanf("%lld", &n);
    for(ll i = 0; i < n; i++)
    {
        scanf("%lld%lld", &x[i], &y[i]);
        cx += x[i];
        cy += y[i];
    }
//    cx /= n;
//    cy /= n;
    cx = x[0];
    cy = y[0];
    for(ll i = 0; i < M; i++)
    {
        f(now);
        now *= Y;
    }
    ld ans = 0;
    for(ll i = 0; i < n; i++)
    {
        ld r = (cx - x[i]) * (cx - x[i]) + (cy - y[i]) * (cy - y[i]);
        ans = max(ans, r);
    }
    printf("%.12Lf", sqrt(ans));
    return 0;
}
