E

传送门

There is a forest that we model as a plane and live nn rare animals. Animal number iihas its lair in the point (xi,yi)(xi,yi). 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=0y=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 nn (1≤n≤1051≤n≤105) — the number of animals. 

Each of the next nn lines contains two integers xixi, yiyi (−107≤xi,yi≤107−107≤xi,yi≤107) — the coordinates of the ii-th animal's lair. It is guaranteed that yi≠0yi≠0. No two lairs coincide.

Output

If the reserve cannot be built, print −1−1. Otherwise print the minimum radius. Your answer will be accepted if absolute or relative error does not exceed 10−610−6.

Formally, let your answer be aa, and the jury's answer be bb. Your answer is considered correct if |a−b|max(1,|b|)≤10−6|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 0.50.5 and the center in (0, 0.5)(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 5858 and the center in (12, 58)(12, 58).

题意：一个圆与X轴相切，问最小的半径，使得圆包含所有给定的点。

题解：显然，所有的点必须都要在y轴的同一边才可以，之后二分半径，求取每一个点在枚举半径的时候的圆心的区间值

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include <set>
#include<queue>
#include<map>
using namespace std;

#define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn =  1e5+5;
const ll mod = 1e9+7;
const double eps = 1e-8;

double x[maxn],y[maxn];
long double maxx = 100000000000000000.0;
int n;
bool check(long double k)
{
    long double l = -maxx,r = maxx;
    long double t;
    for(int i=1;i<=n;i++)
    {
        if(y[i] > k*2)
            return false;
        t = sqrt(y[i]*(2*k-y[i]));
        if(l < x[i]-t)
            l = x[i]-t;
        if(r > x[i]+t)
            r =x[i]+t;
    }
    return l<r;
}
int main()
{
    scanf("%d",&n);
    for(int i = 1;i <= n;i++)
        scanf("%lf %lf",&x[i],&y[i]);
    for(int i=1;i<=n;i++)
    {
        if(y[i] * y[n] < 0)
        {
            puts("-1");
            return 0;
        }
        else
            y[i] = max(y[i],-y[i]);
    }
    long double l = 0,r = maxx,mid;
    for(int i=1;i<=100;i++)
    {
        mid = (l+r)/2.0;
        if(check(mid))
            r = mid;
        else
            l = mid;
    }
    printf("%.8Lf
",mid);
    return 0;
}