Problem: There is a forest that we model as a plane and live n rare animals. Animal number i has its lair in the point (x_{i}, y_{i}). In order to protect them, a decision to build a nature reserve has been made.

The reserve must have a form of a circle containing all lairs. There is also a straight river flowing through the forest. All animals drink from this river, therefore it must have at least one common point with the reserve. On the other hand, ships constantly sail along the river, so the reserve must not have more than one common point with the river.

For convenience, scientists have made a transformation of coordinates so that the river is defined by y = 0. Check whether it is possible to build a reserve, and if possible, find the minimum possible radius of such a reserve.

Input

The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of animals. 

Each of the next n lines contains two integers x_{i}, y_{i} (-10^7 ≤ x_{i}, y_{i} ≤ 10^7) — the coordinates of the i-th animal's lair. It is guaranteed that y_{i} ≠ 0. No two lairs coincide.

Output

If the reserve cannot be built, print -1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.

Examples

Input

1
0 1


Output

0.5

Input

3
0 1
0 2
0 -3


Output

-1


Input

2
0 1
1 1


Output

0.625

Note

In the first sample it is optimal to build the reserve with the radius equal to 0.5 and the center in (0,\ 0.5).

In the second sample it is impossible to build a reserve.

In the third sample it is optimal to build the reserve with the radius equal to 5/8 and the center in (1/2,\ 5/8).

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long gcd(long long a, long long b) {
  return (b == 0LL ? a : gcd(b, a % b));
}
long long exp(long long a, long long b, long long m) {
  if (b == 0LL) return 1LL;
  if (b == 1LL) return mod(a, m);
  long long k = mod(exp(a, b / 2, m), m);
  if (b & 1LL) {
    return mod(a * mod(k * k, m), m);
  } else
    return mod(k * k, m);
}
template <class T>
long long sgn(T x) {
  return (x > 0) - (x < 0);
}
template <class T>
struct Point {
  T x, y;
  explicit Point(T x = 0, T y = 0) : x(x), y(y) {}
  bool operator<(Point p) const { return tie(x, y) < tie(p.x, p.y); }
  bool operator==(Point p) const { return tie(x, y) == tie(p.x, p.y); }
  Point operator+(Point p) const { return Point(x + p.x, y + p.y); }
  Point operator-(Point p) const { return Point(x - p.x, y - p.y); }
  Point operator*(T d) const { return Point(x * d, y * d); }
  Point operator/(T d) const { return Point(x / d, y / d); }
  T dot(Point p) const { return x * p.x + y * p.y; }
  T cross(Point p) const { return x * p.y - y * p.x; }
  T cross(Point a, Point b) const { return (a - *this).cross(b - *this); }
  T dist2() const { return x * x + y * y; }
  double dist() const { return sqrt((double)dist2()); }
  double angle() const { return atan2(y, x); }
  Point unit() const { return *this / dist(); }
  Point perp() const { return Point(-y, x); }
  Point normal() const { return perp().unit(); }
  Point rotate(double a) const {
    return Point(x * cos(a) - y * sin(a), x * sin(a) + y * cos(a));
  }
  friend ostream& operator<<(ostream& os, Point p) {
    return os << "(" << p.x << "," << p.y << ")";
  }
};
long double EPS = 1e-9;
long long n;
vector<Point<long double> > p;
inline long double good_sqrt(long double D) { return D < 0 ? 0 : sqrt(D); }
inline bool bigger(const long double& x, const long double& y) {
  return x > y + EPS;
}
inline bool small(long double& x, long double y) { return y > x + EPS; }
bool ok(long double R) {
  long double l = -1e16 - 1, r = 1e16 + 1;
  for (long long i = 0; i < n; i++) {
    long double b = -2 * p[i].x;
    long double c = p[i].x * p[i].x + p[i].y * p[i].y - 2 * p[i].y * R;
    long double delta = b * b - 4.0 * c;
    if (delta < -EPS) return false;
    long double D = good_sqrt(delta);
    long double x1 = p[i].x - D / 2;
    long double x2 = p[i].x + D / 2;
    l = max(l, x1);
    r = min(r, x2);
  }
  return !(l > r + EPS);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  long long cima = 0, embaixo = 0;
  p.resize(n);
  for (long long i = 0; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    p[i] = Point<long double>(x, y);
    if (y > 0) cima = 1;
    if (y < 0) embaixo = 1;
  }
  if (cima and embaixo) {
    cout << -1 << "\n";
    return 0;
  }
  if (embaixo) {
    for (long long i = 0; i < n; i++) {
      p[i].y = -p[i].y;
    }
  }
  long double l = 0, r = 1e16;
  function<long double(long double, long double)> get_mid;
  if (ok(1)) {
    r = 1;
    get_mid = [](long double l, long double r) { return (l + r) / 2.0; };
  } else {
    l = 1;
    get_mid = [](long double l, long double r) { return sqrt(l * r); };
  }
  for (long long iter = 0; iter < 150; iter++) {
    long double m = get_mid(l, r);
    if (ok(m)) {
      r = m;
    } else
      l = m;
  }
  cout << setprecision(16) << fixed;
  cout << get_mid(l, r) << "\n";
}