Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
#define fi first
#define se second
#define pb push_back

const int N = 102;
int x[N], y[N], n;

ld cost(pair <ld, ld> center) {
    ld ret = 0;
    for (int i = 0; i < n; i++) {
        ld dist = (center.fi - x[i]) * (center.fi - x[i]) + (center.se - y[i]) * (center.se - y[i]);
        ret = max(ret, dist);
    }
    return sqrt(ret);
}

ld ch(ld x_try) {
    ld ly = 0, ry = 1000;
    for (int i = 0; i < 100; i++) {
        ld mid1 = ly + (ry - ly) / 3;
        ld mid2 = ry - (ry - ly) / 3;
        if (cost({x_try, mid1}) < cost({x_try, mid2})) {
            ry = mid2;
        } else {
            ly = mid1;
        }
    }
    return ly;
}

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d %d", &x[i], &y[i]);
    }
    ld lx = 0, rx = 1000;
    for (int i = 0; i < 100; i++) {
        ld mid1 = lx + (rx - lx) / 3;
        ld mid2 = rx - (rx - lx) / 3;
        if (cost({mid1, ch(mid1)}) < cost({mid2, ch(mid2)})) {
            rx = mid2;
        } else {
            lx = mid1;
        }
    }
    printf("%.12Lf", cost({lx, ch(lx)}));
	return 0;
}