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 long long mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = 1e20;
const int maxp = 110;
int dblcmp(double d) {
  if (fabs(d) < eps) return 0;
  return d > eps ? 1 : -1;
}
inline double sqr(double x) { return x * x; }
struct point {
  double x, y;
  point() {}
  point(double _x, double _y) : x(_x), y(_y){};
  void input() { scanf("%lf%lf", &x, &y); }
  void output() { printf("%.2f %.2f\n", x, y); }
  bool operator==(point a) const {
    return dblcmp(a.x - x) == 0 && dblcmp(a.y - y) == 0;
  }
  bool operator<(point a) const {
    return dblcmp(a.x - x) == 0 ? dblcmp(y - a.y) < 0 : x < a.x;
  }
  double len() { return sqrt(x * x + y * y); }
  int len2() { return x * x + y * y + 0.1; }
  double distance(point p) {
    return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));
  }
  point add(point p) { return point(x + p.x, y + p.y); }
  point sub(point p) { return point(x - p.x, y - p.y); }
  point mul(double b) { return point(x * b, y * b); }
  point div(double b) { return point(x / b, y / b); }
  double dot(point p) { return x * p.x + y * p.y; }
  double det(point p) { return x * p.y - y * p.x; }
  double rad(point a, point b) {
    point p = *this;
    return fabs(atan2(fabs(a.sub(p).det(b.sub(p))), a.sub(p).dot(b.sub(p))));
  }
  point trunc(double r) {
    double l = len();
    if (!dblcmp(l)) return *this;
    r /= l;
    return point(x * r, y * r);
  }
  point rotleft() { return point(-y, x); }
  point rotright() { return point(y, -x); }
  point rotate(point p, double angle) {
    point v = this->sub(p);
    double c = cos(angle), s = sin(angle);
    return point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);
  }
};
struct line {
  point a, b;
  line() {}
  line(point _a, point _b) {
    a = _a;
    b = _b;
  }
  bool operator==(line v) { return (a == v.a) && (b == v.b); }
  line(point p, double angle) {
    a = p;
    if (dblcmp(angle - pi / 2) == 0)
      b = a.add(point(0, 1));
    else
      b = a.add(point(1, tan(angle)));
  }
  line(double _a, double _b, double _c) {
    if (dblcmp(_a) == 0)
      a = point(0, -_c / _b), b = point(1, -_c / _b);
    else if (dblcmp(_b) == 0)
      a = point(-_c / _a, 0), b = point(-_c / _a, 1);
    else
      a = point(0, -_c / _b), b = point(1, (-_c - _a) / _b);
  }
  void input() {
    a.input();
    b.input();
  }
  void adjust() {
    if (b < a) swap(a, b);
  }
  double length() { return a.distance(b); }
  double angle() {
    double k = atan2(b.y - a.y, b.x - a.x);
    if (dblcmp(k) < 0) k += pi;
    if (dblcmp(k - pi) == 0) k -= pi;
    return k;
  }
  int relation(point p) {
    int c = dblcmp(p.sub(a).det(b.sub(a)));
    if (c < 0) return 1;
    if (c > 0) return 2;
    return 3;
  }
  bool pointonseg(point p) {
    return dblcmp(p.sub(a).det(b.sub(a))) == 0 &&
           dblcmp(p.sub(a).dot(p.sub(b))) <= 0;
  }
  bool parallel(line v) { return dblcmp(b.sub(a).det(v.b.sub(v.a))) == 0; }
  int segcrossseg(line v) {
    int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));
    int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));
    int d3 = dblcmp(v.b.sub(v.a).det(a.sub(v.a)));
    int d4 = dblcmp(v.b.sub(v.a).det(b.sub(v.a)));
    if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;
    return (d1 == 0 && dblcmp(v.a.sub(a).dot(v.a.sub(b))) <= 0 ||
            d2 == 0 && dblcmp(v.b.sub(a).dot(v.b.sub(b))) <= 0 ||
            d3 == 0 && dblcmp(a.sub(v.a).dot(a.sub(v.b))) <= 0 ||
            d4 == 0 && dblcmp(b.sub(v.a).dot(b.sub(v.b))) <= 0);
  }
  int linecrossseg(line v) {
    int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));
    int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));
    if ((d1 ^ d2) == -2) return 2;
    return (d1 == 0 || d2 == 0);
  }
  int linecrossline(line v) {
    if ((*this).parallel(v)) {
      return v.relation(a) == 3;
    }
    return 2;
  }
  point crosspoint(line v) {
    double a1 = v.b.sub(v.a).det(a.sub(v.a));
    double a2 = v.b.sub(v.a).det(b.sub(v.a));
    return point((a.x * a2 - b.x * a1) / (a2 - a1),
                 (a.y * a2 - b.y * a1) / (a2 - a1));
  }
  double dispointtoline(point p) {
    return fabs(p.sub(a).det(b.sub(a))) / length();
  }
  double dispointtoseg(point p) {
    if (dblcmp(p.sub(b).dot(a.sub(b))) < 0 ||
        dblcmp(p.sub(a).dot(b.sub(a))) < 0) {
      return min(p.distance(a), p.distance(b));
    }
    return dispointtoline(p);
  }
  point lineprog(point p) {
    return a.add(b.sub(a).mul(b.sub(a).dot(p.sub(a)) / b.sub(a).len2()));
  }
  point symmetrypoint(point p) {
    point q = lineprog(p);
    return point(2 * q.x - p.x, 2 * q.y - p.y);
  }
};
line midl(point a, point b) {
  point u = a.add(b).div(2), v = b.sub(a).rotleft();
  return line(u, u.add(v));
}
point circle(point a, point b, point c) {
  return line(a.add(b).div(2), a.add(b).div(2).add(b.sub(a).rotleft()))
      .crosspoint(
          line(c.add(b).div(2), c.add(b).div(2).add(b.sub(c).rotleft())));
}
const int N = 110;
int n;
point p[N];
double ret = -1;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) p[i].input();
  for (int u = 0; u < n; u++)
    for (int v = u + 1; v < n; v++)
      for (int w = v + 1; w < n; w++) {
        int a = p[u].sub(p[v]).len2(), b = p[v].sub(p[w]).len2(),
            c = p[u].sub(p[w]).len2();
        if (a + b >= c && a + c >= b && b + c >= a) {
          point o = circle(p[u], p[v], p[w]);
          double r = p[u].distance(o);
          int c0 = 0, c1 = 0;
          for (int i = 0; i < n; i++)
            if (i != u && i != v && i != w) {
              double dr = p[i].distance(o);
              if (dr < r - eps) {
                c0++;
                break;
              } else if (fabs(dr - r) <= eps) {
                c1++;
              }
            }
          if (c0) continue;
          if (c1 || (a + b > c && a + c > b && b + c > a)) ret = max(ret, r);
        }
      }
  printf("%.10f\n", ret);
}