Problem:
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Circles Intersection

You are given circle $A$ with radius $r_a$ and with central coordinate $(x_a, y_a)$ and circle $B$ with radius $r_b$ and with central coordinate $(x_b, y_b)$.

Write a program which prints:

"2" if $B$ is in $A$,

"-2" if $A$ is in $B$,

"1" if circumference of $A$ and $B$ intersect, and

"0" if $A$ and $B$ do not overlap.

You may assume that $A$ and $B$ are not identical.

Input

The input consists of multiple datasets. The first line consists of an integer $N$ ($N \leq 50$), the number of datasets. There will be $N$ lines where each line represents each dataset. Each data set consists of real numbers:

$x_a$ $y_a$ $r_a$ $x_b$ $y_b$ $r_b$

Output

For each dataset, print 2, -2, 1, or 0 in a line.

Sample Input

2
0.0 0.0 5.0 0.0 0.0 4.0
0.0 0.0 2.0 4.1 0.0 2.0

Output for the Sample Input

2
0

Solution in JAVA (judge verdict: Accepted)

import java.io.*;

public class Main{
	public static void main(String[] args){
		BufferedReader r=new BufferedReader(new InputStreamReader(System.in));
		String s,u;
		String[] a;
		double xa,ya,ra,xb,yb,rb,x,y,l1,l2;
		double l;
		int ans=0;
		int t;
		try{
				while((s=r.readLine()) !=null){
						t=Integer.parseInt(s);
						for(int i=0;i<t;i++){
							u=r.readLine();
							a=u.split(" ");
							double[] b=new double[a.length];
							for(int j=0;j<a.length;j++){
								b[j]=Double.parseDouble(a[j]);
							}
							xa=b[0];
							ya=b[1];
							ra=b[2];
							xb=b[3];
							yb=b[4];
							rb=b[5];
							if(xa>xb){
								x=xa-xb;
							}else{
								x=xb-xa;
							}
							if(ya>yb){
								y=ya-yb;
							}else{
								y=yb-ya;
							}
							l=Math.sqrt(x*x+y*y);
							l1=ra+rb;
							if(ra>rb){
								l2=ra-rb;
							}else{
								l2=rb-ra;
							}
							if(l<l2){
								ans=2;
								if(rb>ra){
									ans *=-1;
								}
							}else{ 
								if(l2<=l &&l<=l1){
										ans=1;
							    }else{
										ans=0;
							    }
							}
							System.out.println(ans);
						}
				}
		}catch(IOException e){
				System.out.println(e);
		}
	}
}