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;
template <class It>
void dptab(It b, It e, const char *f = "%d ") {
  if (1) {
    for (It it = b; it != e; ++it)
      if (1) {
        printf("\033[31m");
        printf(f, *it);
        printf("\033[0m");
      };
    if (1) {
      printf("\033[31m");
      printf("\n");
      printf("\033[0m");
    };
  }
}
const double EPS = 1e-6;
struct pkt {
  double x, y;
  explicit pkt(double x = 0, double y = 0) : x(x), y(y) {}
  double dist(const pkt &p) { return sqrt(pow(x - p.x, 2) + pow(y - p.y, 2)); }
  pkt operator+(const pkt &p) { return pkt(x + p.x, y + p.y); }
  pkt operator/(double s) { return pkt(x / s, y / s); }
};
struct line {
  double A, B, C;
  explicit line(double A = 0, double B = 0, double C = 0) : A(A), B(B), C(C) {}
  explicit line(const pkt &a, const pkt &b) {
    A = a.y - b.y;
    B = b.x - a.x;
    C = -a.x * A - a.y * B;
  }
  line perpendicular(const pkt &p) { return line(-B, A, B * p.x - A * p.y); }
  pkt intersection_point(const line &l) {
    return pkt(l.C * B - C * l.B, l.A * C - A * l.C) / (A * l.B - l.A * B);
  }
};
struct segment {
  pkt a, b;
  explicit segment(const pkt &a, const pkt &b) : a(a), b(b) {}
  line bisector() { return line(a, b).perpendicular((a + b) / 2); }
  int point_side(const pkt &p) {
    double res =
        a.x * b.y + b.x * p.y + p.x * a.y - (b.x * a.y + p.x * b.y + a.x * p.y);
    if (abs(res) < EPS)
      return 0;
    else if (res < 0)
      return -1;
    else
      return 1;
  }
};
struct triangle {
  pkt a, b, c;
  explicit triangle(const pkt &a, const pkt &b, const pkt &c)
      : a(a), b(b), c(c) {}
  bool inside(pkt &p) {
    int side1, side2, side3;
    side1 = segment(a, b).point_side(p);
    side2 = segment(b, c).point_side(p);
    side3 = segment(c, a).point_side(p);
    return side1 == side2 && side2 == side3;
  }
  bool on_edge(pkt &p) {
    int side1, side2, side3;
    side1 = segment(a, b).point_side(p);
    side2 = segment(b, c).point_side(p);
    side3 = segment(c, a).point_side(p);
    return side1 == 0 && side2 == side3 || side2 == 0 && side1 == side3 ||
           side3 == 0 && side2 == side1 || side1 == 0 && side2 == 0 ||
           side3 == 0 && side2 == 0 || side1 == 0 && side3 == 0;
  }
  pkt circumcenter() {
    line b1 = segment(a, b).bisector();
    line b2 = segment(b, c).bisector();
    return b1.intersection_point(b2);
  }
};
ostream &operator<<(ostream &out, pkt &P) {
  return out << "(" << P.x << "," << P.y << ")";
}
ostream &operator<<(ostream &out, line &l) {
  return out << "A=" << l.A << "\tB=" << l.B << "\tC=" << l.C;
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  scanf("%d", &n);
  vector<pkt> v;
  for (int i = (0); i < (n); ++i) {
    pkt p;
    scanf("%lf", &p.x);
    scanf("%lf", &p.y);
    v.push_back(p);
  }
  double last_hole = -1.0;
  for (int i = (0); i < (n); ++i)
    for (int j = (i + 1); j < (n); ++j)
      for (int k = (j + 1); k < (n); ++k) {
        pkt p1 = v[i];
        pkt p2 = v[j];
        pkt p3 = v[k];
        triangle t(p1, p2, p3);
        if (segment(p1, p2).point_side(p3) == 0) continue;
        pkt center = t.circumcenter();
        if (!t.inside(center) && !t.on_edge(center)) {
          continue;
        }
        bool right = false;
        if (t.on_edge(center)) {
          right = true;
        }
        bool points_ok = true;
        for (int l = (0); l < (n); ++l) {
          if (l == i || l == j || l == k) continue;
          pkt p = v[l];
          if (t.inside(p)) {
            points_ok = false;
            break;
          }
          if (right && abs(p.dist(center) - p1.dist(center)) < EPS) {
            pkt opposite;
            segment s3(p1, p2), s2(p1, p3), s1(p2, p3);
            if (s3.point_side(center) == 0) {
              if (s3.point_side(p) != s3.point_side(p3)) right = false;
            } else if (s2.point_side(center) == 0) {
              if (s2.point_side(p) != s2.point_side(p2)) right = false;
            } else {
              if (s1.point_side(p) != s1.point_side(p1)) right = false;
            }
          }
          if ((p.dist(center) + EPS) < p1.dist(center)) {
            points_ok = false;
            break;
          }
        }
        if (points_ok && !right) {
          double distance = p1.dist(center);
          last_hole = max(last_hole, distance);
        }
      }
  if (last_hole == -1)
    printf("-1\n");
  else
    printf("%lf\n", last_hole);
  return 0;
}