Codeforces #514 D. Nature Reserve（几何 最小圆覆盖，三分）

D. Nature Reserve

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a forest that we model as a plane and live nn rare animals. Animal number ii has 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

Copy

1
0 1

output

Copy

0.5

input

Copy

3
0 1
0 2
0 -3

output

Copy

-1

input

Copy

2
0 1
1 1

output

Copy

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).

【体验感】

当机会摆在眼前，我却没有珍惜它，等到被eps卡炸，我菜追悔莫及。long double的有效位数18位，能解方程的问题我非要二分查找，这题答案最大能到1e14，还剩4位能给eps用，eps最小只能1e-4了，再小就跑死了。这题答案要求误差1e-6。把自己搞死了。   为什么要二分？为什么要二分？明明解一解方程就能求的事情，懒，是懒，懒到不想写方程，懒到二分暴力一下。

求与x轴相切的最小圆覆盖。

【题意】

平面上给出一些点（注意数据范围），求一个最小的圆的半径，满足此圆覆盖所有点并且与x轴相切！

【分析】

经验算，圆的半径最坏情况能到1e14，导致了我的【体验感】。

易知，这个圆的圆心坐标x，越靠中间，其可行圆的半径越小，因此可以用三分找到这个x坐标。但是三分时需要求的当前x坐标下的半径是多少？   

下面求给定x坐标后的半径：对于任意一点(x1,y1)，设最小圆圆心(x,y)，半径r，则满足：r==y && r*r==(y-y1)*(y-y1)+(x-x1)*(x-x1);解方程即可得到精确的y，也就是r！（而我当时傻的一批，非要用二分求这个r）

【代码】

/****
***author: winter2121
****/
#include
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define read(i) scanf("%d",&i)
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int MAX=1e6+5;
const int INF=0x3f3f3f3f;
const double eps=1e-8;

struct point{
    double x,y;
}p[MAX];
int n;
long double calcu(long double x) //求最小圆半径
{
    long double r=0;
    for(int i=0;i0)flag|=1;
        if(p[i].y<0)flag|=2;
        p[i].y=fabs(p[i].y);
    }
    if(flag==3) return puts("-1"),0;
    long double l=-1e7,r=1e7,lx,rx,dx; //枚举圆心
    while(r-l>eps)
    {
        dx=(r-l)/3.0;
        lx=l+dx;
        rx=r-dx;
        if(calcu(lx)-calcu(rx)<0)r=rx;
        else l=lx;
    }
    printf("%.8f\n",(double)calcu(r));
    return 0;
}