Problem: Luyi has n circles on the plane. The i-th circle is centered at (xi, yi). At the time zero circles start to grow simultaneously. In other words, the radius of each circle at time t (t > 0) is equal to t. The circles are drawn as black discs on an infinite white plane. So at each moment the plane consists of several black and white regions. Note that the circles may overlap while growing.

<image>

We define a hole as a closed, connected white region. For instance, the figure contains two holes shown by red border. During growing some holes may be created and it is easy to see that each created hole will disappear eventually. Luyi asks you to find moment of time such that the last hole disappears. In other words, you should find the first moment such that no hole can be seen after that.

Input

The first line of the input contains integer n (1 ≤ n ≤ 100). Each of the next n lines contains two integers xi and yi ( - 104 ≤ xi, yi ≤ 104), indicating the location of i-th circle.

It's guaranteed that no two circles are centered at the same point.

Output

Print the moment where the last hole disappears. If there exists no moment in which we can find holes print -1.

The answer will be considered correct if the absolute or relative error does not exceed 10 - 4.

Examples

Input

3
0 0
1 1
2 2


Output

-1


Input

4
0 0
0 2
2 2
2 0


Output

1.414214


Input

4
0 1
0 -1
-2 0
4 0


Output

2.125000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 111;
const long double Eps = 1e-6;
int sgn(long double a, long double b = 0) {
  a -= b;
  return (a > Eps) - (a < -Eps);
}
const long double Pi = 4 * atan(1);
struct Point {
  long double x, y;
  Point operator-(Point p) { return {x - p.x, y - p.y}; }
  Point operator+(Point p) { return {x + p.x, y + p.y}; }
  Point operator/(long double d) { return {x / d, y / d}; }
  bool operator<(Point p) const {
    if (sgn(x, p.x) == 0) return sgn(y, p.y) < 0;
    return sgn(x, p.x) < 0;
  }
  Point rot(long double d) {
    long double rx = x * cos(d) - y * sin(d);
    long double ry = x * sin(d) + y * cos(d);
    return {rx, ry};
  }
  long double len2() { return x * x + y * y; }
  long double len() { return sqrt(len2()); }
  long double dot(Point p) { return x * p.x + y * p.y; }
  long double cross(Point p) { return x * p.y - y * p.x; }
} p[N];
long double xmul(Point a, Point b, Point c) { return (b - a).cross(c - a); }
Point inter(Point p1, Point p2, Point q1, Point q2) {
  long double u = xmul(p1, p2, q1);
  long double v = xmul(p2, p1, q2);
  long double rx = (q1.x * v + q2.x * u) / (v + u);
  long double ry = (q1.y * v + q2.y * u) / (v + u);
  return {rx, ry};
}
class Cmp {
 public:
  bool operator()(long double a, long double b) const { return sgn(a, b) < 0; }
};
int main() {
  ios ::sync_with_stdio(false);
  int n;
  cin >> n;
  set<Point> st;
  for (int i = 0; i < n; i++) {
    cin >> p[i].x >> p[i].y;
    st.insert(p[i]);
  }
  bool found = false;
  long double ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      for (int k = j + 1; k < n; k++) {
        if (sgn((p[j] - p[i]).dot(p[k] - p[i])) > 0 &&
            sgn((p[k] - p[j]).dot(p[i] - p[j])) > 0 &&
            sgn((p[i] - p[k]).dot(p[j] - p[k])) > 0) {
          Point mij = (p[i] + p[j]) / 2;
          Point dij = (p[i] - p[j]).rot(Pi / 2);
          Point mjk = (p[j] + p[k]) / 2;
          Point djk = (p[j] - p[k]).rot(Pi / 2);
          Point o = inter(mij, mij + dij, mjk, mjk + djk);
          bool flag = true;
          for (int r = 0; r < n; r++)
            if (sgn((p[r] - o).len2(), (p[i] - o).len2()) < 0) flag = false;
          if (flag) {
            found = true;
            ans = max(ans, (p[i] - o).len());
          }
        }
      }
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) {
      Point mij = (p[i] + p[j]) / 2;
      Point dij = (p[j] - p[i]) / 2;
      Point pk = mij + dij.rot(Pi / 2);
      Point ps = mij - dij.rot(Pi / 2);
      if (st.find(pk) != st.end() && st.find(ps) != st.end()) {
        Point &o = mij;
        bool flag = true;
        for (int s = 0; s < n; s++)
          if (sgn((p[s] - o).len2(), (p[i] - o).len2()) < 0) flag = false;
        if (flag) {
          found = true;
          ans = max(ans, (p[i] - o).len());
        }
      }
    }
  if (!found)
    cout << -1 << '\n';
  else
    cout << setprecision(6) << ans << '\n';
  return 0;
}