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>
#pragma warning(disable : 4996)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("-ffloat-store")
using namespace std;
clock_t time_p = clock();
void aryanc403() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
class CMP {
 public:
  bool operator()(long long int a, long long int b) { return !(a <= b); }
};
void add(map<long long int, long long int> &m, long long int x) {
  map<long long int, long long int>::iterator jt;
  jt = m.find(x);
  if (jt == m.end())
    m.insert(make_pair(x, 1));
  else
    jt->second++;
}
void del(map<long long int, long long int> &m, long long int x) {
  map<long long int, long long int>::iterator jt;
  jt = m.find(x);
  if (jt->second == 1)
    m.erase(jt);
  else
    jt->second--;
}
bool cmp(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return a.first > b.first || (a.first == b.first && a.second < b.second);
}
const long long int INF = 0xFFFFFFFFFFFFFFFL;
const long long int mod = 1000000007L;
long long int T, n, i, m, j, k, in, cnt, l, r;
vector<pair<long long int, long long int> > a;
vector<long long int>::iterator it;
string s;
const long double esp = 1e-9;
const long long int itr = 1000;
bool check(long double r) {
  long double x = 0, y = 0, c;
  x -= 1e10;
  y += 1e10;
  for (auto &pt : a) {
    if (2 * r - pt.second < -esp) return false;
    c = (2 * r - pt.second) * pt.second;
    c = sqrt(c);
    x = max(x, pt.first - c);
    y = min(y, pt.first + c);
  }
  if (x - y > esp) return false;
  return true;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  {
    bool fl = false;
    cin >> n;
    long long int x, y;
    a.clear();
    a.reserve(n);
    for (i = 0; i < (n); ++i) {
      cin >> x >> y;
      if (i == 0) {
        if (y > 0)
          fl = false;
        else {
          y = -y;
          fl = true;
        }
      } else {
        if ((fl && y > 0) || ((!fl) && y < 0)) {
          cout << -1;
          return 0;
        }
        if (fl) y = -y;
      }
      a.push_back(make_pair(x, y));
    }
    long double l = 0, r = 0, mid;
    r += 1e16;
    long long int i = 0;
    while (i < itr && fabs(r - l) > esp) {
      mid = (l + r) / 2;
      if (check(mid))
        r = mid;
      else
        l = mid;
      i++;
    }
    cout << std::fixed << std::setprecision(30) << l;
  }
  aryanc403();
  return 0;
}