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;
const int INF = 1e9;
const long long INFLL = 1e18 + 1;
const int MAX = 200001;
const long long MOD = 1000000007;
const int SZ = 300100;
const double eps = 0.000000006;
const double PI = 3.14159265358979323846264338327;
long long inq(long long k, long long q, long long mod) {
  if (q == 0) return 1;
  long long l = inq(k, q / 2, mod);
  if (q % 2 == 0)
    return l * l % mod;
  else
    return l * l % mod * k % mod;
}
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long cubr(long long a) {
  long long l = -1, r = 1e6 + 2;
  while (l < r - 1) {
    long long mid = (l + r) / 2;
    if (mid * mid * mid > a)
      r = mid;
    else
      l = mid;
  }
  return l;
}
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long min(long long a, long long b) { return -1 * max(-a, -b); }
long long possible(long long q) {
  if (q == INF) return -1;
  return q;
}
bool correct(int x, int xx) {
  if (x < 0) return 0;
  if (x >= xx) return 0;
  return 1;
}
long long dsumm(long long x, long long k) {
  long long y = 1;
  long long z = 0;
  for (int i = 0; y < 1e18; i++) {
    z += x / y % k;
    y *= k;
  }
  return z;
}
long long dcount(long long x) {
  long long y = 1;
  long long z = 0;
  int c[100];
  for (int i = 0; i < 10; i++) c[i] = 0;
  for (int i = 0; x > 0; i++) {
    if (c[x / y % 10] == 0) z++;
    c[x / y % 10] = 1;
    x /= 10;
  }
  return z;
}
long long lg10(long long x) {
  if (10000 <= x && x < 100000) return 5;
  if (x == 0) return 0;
  return lg10(x / 10) + 1;
}
long long g(long long x, long long mod) {
  if (x == 0) return 0;
  return x / mod + g(x / mod, mod);
}
bool is_digit(char c) { return ('0' <= c && c <= '9'); }
long long n, m, k, L, b;
pair<double, double> p[500000];
double dist(double x1, double y1, double x2, double y2) {
  return sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));
}
bool c2(double x, double r) {
  for (int i = 0; i < n; i++) {
    if (dist(x, r, p[i].first, p[i].second) > r + eps) return 0;
  }
  return 1;
}
bool c(double r) {
  double minimum = 1e15;
  for (int i = 0; i < n; i++) {
    double xi = p[i].first;
    double yi = p[i].second;
    double optimal = xi + sqrt(r * r - (r - yi) * (r - yi));
    if (optimal < minimum) minimum = optimal;
  }
  return c2(minimum, r);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  long long pos = 0, neg = 0;
  for (int i = 0; i < n; i++) {
    cin >> p[i].first >> p[i].second;
    if (p[i].second > 0)
      pos++;
    else {
      neg++;
      p[i].second *= -1;
    }
  }
  if (pos > 0 && neg > 0) {
    cout << -1;
    return 0;
  }
  long long it = 69;
  for (int(i) = 0; (i) != (n); i++) {
  }
  double l = 0, r = 1e15;
  while (it >= 0) {
    double mid = (l + r) / 2;
    if (c(mid) == 1)
      r = mid;
    else
      l = mid;
    it--;
  }
  printf("%.10f", r);
}