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>
void Get(int &T) {
  char C;
  bool F = 0;
  for (; C = getchar(), C < '0' || C > '9';)
    if (C == '-') F = 1;
  for (T = C - '0'; C = getchar(), C >= '0' && C <= '9'; T = T * 10 + C - '0')
    ;
  F && (T = -T);
}
int N;
int X[105];
int Y[105];
void Init() {
  Get(N);
  for (int i = 1; i <= N; i++) {
    Get(X[i]), Get(Y[i]);
  }
}
long long Dis2(int A, int B) {
  long long DX = X[A] - X[B];
  long long DY = Y[A] - Y[B];
  long long Ans = DX * DX + DY * DY;
  return Ans;
}
double Dis2(int A, double XB, double YB) {
  double DX = X[A] - XB;
  double DY = Y[A] - YB;
  double Ans = DX * DX + DY * DY;
  return Ans;
}
double CP(double X1, double Y1, double X2, double Y2) {
  return X1 * Y2 - X2 * Y1;
}
int Cp(int P0, int P1, int P2) {
  return ((long long)(X[P1] - X[P0])) * ((long long)(Y[P2] - Y[P0])) -
         ((long long)(X[P2] - X[P0])) * ((long long)(Y[P1] - Y[P0]));
}
double Ans;
void Judge(double X, double Y, double Time) {
  if (Ans >= Time) return;
  for (int i = 1; i <= N; i++)
    if (Dis2(i, X, Y) + (1e-8) < Time) {
      return;
    }
  Ans = Time;
}
void Cir(int A, int B, int C) {
  double X1 = X[A] + X[B];
  X1 *= 0.5;
  double Y1 = Y[A] + Y[B];
  Y1 *= .5;
  double X2 = Y[A] - Y[B] + X1;
  double Y2 = X[B] - X[A] + Y1;
  double X3 = X[C] + X[B];
  X3 *= 0.5;
  double Y3 = Y[C] + Y[B];
  Y3 *= .5;
  double X4 = Y[C] - Y[B] + X3;
  double Y4 = X[B] - X[C] + Y3;
  double S1 = CP(X2 - X1, Y2 - Y1, X3 - X1, Y3 - Y1);
  double S2 = -CP(X2 - X1, Y2 - Y1, X4 - X1, Y4 - Y1);
  double X0 = S2 * X3 + S1 * X4;
  X0 /= S1 + S2;
  double Y0 = S2 * Y3 + S1 * Y4;
  Y0 /= S1 + S2;
  double Time = Dis2(A, X0, Y0);
  Judge(X0, Y0, Time);
}
void Work() {
  for (int i = 1; i <= N; i++)
    for (int j = i + 1; j <= N; j++) {
      int Cnt1 = 0;
      int Cnt2 = 0;
      for (int k = 1; k <= N; k++) {
        if (i == j || j == k || k == i) continue;
        long long A2 = Dis2(i, j);
        long long B2 = Dis2(k, j);
        long long C2 = Dis2(i, k);
        if (A2 + B2 > C2 && A2 + C2 > B2 && B2 + C2 > A2) {
          Cir(i, j, k);
        }
        if (B2 + C2 == A2 && Cp(i, j, k) > 0) {
          Cnt1++;
        }
        if (B2 + C2 == A2 && Cp(i, j, k) < 0) {
          Cnt2++;
        }
      }
      if (Cnt1 > 0 && Cnt2 > 0) {
        Judge((X[i] + X[j]) * .5, (Y[i] + Y[j]) * .5, Dis2(i, j) * .25);
      }
    }
}
void Output() {
  if (Ans == 0)
    puts("-1");
  else {
    Ans = sqrt(Ans);
    printf("%.8lf\n", Ans);
  }
}
int main() {
  Init();
  Work();
  Output();
  return 0;
}