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 MOD = 998244353;
const int N = 200005;
const double PI = 4 * atan(1);
const long long MAX = 9223372036854775807;
int dx[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, 1, -1, 1, 1, -1, -1};
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
bool issquare(long long w) { return trunc(sqrt(w)) * trunc(sqrt(w)) == w; }
bool isprime(long long u) {
  for (long long i = 2; i <= (int)sqrt(u); i++) {
    if (u % i == 0) return 0;
  }
  return 1;
}
long long mod(long long to_mod, long long MOD) {
  to_mod %= MOD;
  while (to_mod < 0) to_mod += MOD;
  return to_mod % MOD;
}
long long moduloMultiplication(long long a, long long b, long long mod) {
  long long res = 0;
  a %= mod;
  while (b) {
    if (b & 1) res = (res + a) % mod;
    a = (2 * a) % mod;
    b >>= 1;
  }
  return res;
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long _sieve_size;
bitset<15000100> bs;
vector<long long> primes;
void sieve(long long upperbound) {
  _sieve_size = upperbound + 1;
  bs.set();
  bs[0] = bs[1] = 0;
  for (long long i = 2; i <= _sieve_size; i++)
    if (bs[i]) {
      for (long long j = i * i; j <= _sieve_size; j += i) bs[j] = 0;
      primes.push_back(i);
    }
}
long double n, m, x, y, t, k, z, w, l, r, ans, a;
vector<pair<long double, long double> > v;
bool pos = false, neg = false;
bool check(long double r) {
  long double trollleft = -1e17;
  long double trollright = 1e17;
  for (int i = 0; i < n; i++) {
    long double easy = sqrt(r * r - (r - v[i].second) * (r - v[i].second));
    long double le = v[i].first - easy;
    long double ri = v[i].first + easy;
    trollleft = max(trollleft, le);
    trollright = min(trollright, ri);
    if (trollright < trollleft) return false;
  }
  return true;
}
int main() {
  cin >> n;
  l = 1e9;
  r = -1e9;
  long double lef = 0, righ = 1e12;
  for (int i = 0; i < n; i++) {
    cin >> x >> y;
    l = min(l, x / 100.0);
    r = max(r, x / 100.0);
    if (y < 0)
      neg = true;
    else
      pos = true;
    lef = max(lef, abs(y) / 200.0);
    v.push_back(make_pair(x / 100.0, abs(y) / 100.0));
  }
  if (n == 1) {
    printf("%.20Lf", v[0].second * 50.0);
    return 0;
  }
  if (neg && pos) {
    cout << -1;
    return 0;
  }
  while (abs(lef - righ) > (1e-7) / 2) {
    long double mid = (lef + righ) / 2;
    if (check(mid)) {
      righ = mid;
    } else {
      lef = mid;
    }
  }
  long double mid = (lef + righ) / 2;
  if (mid != righ) {
    if (check(mid)) {
      righ = mid;
    } else {
      lef = mid;
    }
  }
  printf("%.8Lf", (lef + righ) * 50.0);
  return 0;
}