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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class A {

	
static double EPS=1e-9;
	
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner();
		PrintWriter out=new PrintWriter(System.out);
		int n=sc.nextInt();
		Centre[]centres=new Centre[n];
		boolean pos=false,neg=false;
		for(int i=0;i<n;i++)
		{
			centres[i]=new Centre(sc.nextInt(),sc.nextInt());
			if(centres[i].y>0)
				pos=true;
			else
				neg=true;
		}
		if(pos && neg) {
			System.out.println(-1);
			return;
		}
		double lo=0,hi=1e16;
		for(Centre centre:centres) 
		{
			if(neg)
				centre.y*=-1;
			lo=Math.max(lo, centre.y);
			
		}
		lo/=2;
		double ans=-1;
		
		for(int i=0;i<100;i++)
		{
			double R=(lo+hi)/2;
			
			double l=-1e18,r=1e18;
			for(Centre centre:centres)
			{
				double []x=centre.getInersection(R);
				l=Math.max(l, x[0]);
				r=Math.min(r, x[1]);
			}
			if(l<r+EPS) {
				hi=R;
				ans=R;
			}
			else
				lo=R;
		}
		out.printf("%.9f\n",ans);
		out.close();
		
	}
	static class Centre
	{
		int x,y;
		
		Centre(int a,int b){
			x=a;
			y=b;
		
		}
		// returns {l,r} which are the left and right border of the circle intersection with the line y=R
		double[] getInersection(double r) throws Exception
		{
			
			
			double a=Math.abs(r-y);
		
			double z=Math.sqrt(r-a)*Math.sqrt(r+a);
			return new double[] {x-z,x+z};
			
			
		}
	}
	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
	}

}