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 long long maxn = 500005;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const long long MOD = 100000007;
const double eps = 1e-10;
long long qpow(long long a, long long b) {
  long long tmp = a % MOD, ans = 1;
  while (b) {
    if (b & 1) {
      ans *= tmp, ans %= MOD;
    }
    tmp *= tmp, tmp %= MOD, b >>= 1;
  }
  return ans;
}
long long lowbit(long long x) { return x & -x; }
long long max(long long a, long long b) { return a > b ? a : b; }
long long min(long long a, long long b) { return a < b ? a : b; }
long long mmax(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
long long mmin(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
void mod(long long &a) {
  a += MOD;
  a %= MOD;
}
long long ll(long long p) { return p << 1; }
long long rr(long long p) { return p << 1 | 1; }
long long mm(long long l, long long r) { return (l + r) / 2; }
long long lg(long long x) {
  if (x == 0) return 1;
  return (long long)log2(x) + 1;
}
bool smleql(double a, double b) {
  if (a < b || fabs(a - b) <= eps) return true;
  return false;
}
bool bigeql(double a, double b) {
  if (a > b || fabs(a - b) <= eps) return true;
  return false;
}
bool eql(double a, double b) {
  if (fabs(a - b) < eps) return 1;
  return 0;
}
double len(double a, double b, double c, double d) {
  return sqrt((a - c) * (a - c) + (b - d) * (b - d));
}
bool isp(long long x) {
  if (x == 1) return false;
  if (x == 2) return true;
  for (long long i = 2; i * i <= x; ++i)
    if (x % i == 0) return false;
  return true;
}
inline long long read() {
  char ch = getchar();
  long long s = 0, w = 1;
  while (ch < 48 || ch > 57) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= 48 && ch <= 57) {
    s = (s << 1) + (s << 3) + ch - 48;
    ch = getchar();
  }
  return s * w;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (!(a & 1) && !(b & 1))
    return gcd(a >> 1, b >> 1) << 1;
  else if (!(b & 1))
    return gcd(a, b >> 1);
  else if (!(a & 1))
    return gcd(a >> 1, b);
  else
    return gcd(abs(a - b), min(a, b));
}
long long lcm(long long x, long long y) { return x * y / gcd(x, y); }
long long n;
struct point {
  double x, y;
  void read() { cin >> x >> y; }
} p[maxn];
bool chk(double R) {
  double l = -inf, r = inf;
  for (long long i = 1; i <= n; ++i) {
    double x = p[i].x, y = p[i].y;
    if (2 * y * R - y * y < 0.0) return 0;
    double len = sqrt(2 * y * R - y * y);
    if (x - len > r || x + len < l) return 0;
    l = max(l, x - len), r = min(r, x + len);
  }
  return 1;
}
double half(double l, double r) {
  long long cnt = 1000;
  while (cnt--) {
    double m = (l + r) / 2.0;
    if (chk(m))
      r = m;
    else
      l = m;
  }
  return l;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  bool neg = 0, pos = 0;
  for (long long i = 1; i <= n; ++i) {
    p[i].read();
    if (p[i].y > 0.0)
      pos = 1;
    else
      neg = 1;
  }
  if (pos && neg) {
    cout << -1;
    return 0;
  }
  for (long long i = 1; i <= n; ++i) {
    p[i].y = fabs(p[i].y);
  }
  double ans = half(0, 1e17);
  cout << ans;
  return 0;
}