Problem: For given two segments s1 and s2, print the distance between them.

s1 is formed by end points p0 and p1, and s2 is formed by end points p2 and p3.

Constraints

* 1 ≤ q ≤ 1000
* -10000 ≤ xpi, ypi ≤ 10000
* p0 ≠ p1 and p2 ≠ p3.

Input

The entire input looks like:


q (the number of queries)
1st query
2nd query
...
qth query


Each query consists of integer coordinates of end points of s1 and s2 in the following format:


xp0 yp0 xp1 yp1 xp2 yp2 xp3 yp3


Output

For each query, print the distance. The output values should be in a decimal fraction with an error less than 0.00000001.

Example

Input

3
0 0 1 0 0 1 1 1
0 0 1 0 2 1 1 2
-1 0 1 0 0 1 0 -1


Output

1.0000000000
1.4142135624
0.0000000000

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define EPS 1e-10
#define equals(a,b) (fabs( (a) - (b) )< EPS )
typedef struct point{
	double x,y;
	point(){};
	point(double x ,double y):x(x),y(y){};
	point operator + (point &p){ return point(x+p.x,y+p.y);	}
	point operator - (point &p){ return point(x-p.x,y-p.y);	}
	point operator * (point &p){ return point(x*p.x-y*p.y,x*p.y+y*p.x) ;}
	point operator * (double a){ return point(x*a,y*a);	}
	point operator / (double a){ return point(x/a,y/a);	}
	
	double abs() { return sqrt(norm()); }
	double norm() { return x*x+y*y; }
	
	bool operator < (const point &p) const { return x!=p.x ? x<p.x : y<p.y; }
	bool operator == (const point &p) const { return fabs(x-p.x)<EPS && fabs(y-p.y)<EPS; }
}point;
double abs(point a){return a.abs();}
double norm(point a){return a.norm();}

typedef complex<double> C;
typedef struct { point s,e;} line;

C convert(point a){ return C(a.x,a.y); }
point convert( C a){ return point(a.real(),a.imag() );}
double dot(point a,point b){ return a.x*b.x+a.y*b.y ; } //内積　a・b
double cross(point a,point b){ return a.x*b.y - a.y*b.x ; }//外積(z成分)　a×b
point vec(line l){return l.e-l.s;}
line make(point s,point e){
	line res; res.s=s; res.e=e;
	return res;
}
point make(){ 
	double x,y; cin>>x>>y;
       	return point(x,y);
}
//直交
bool isorthogonal(point a,point b){ return equals(dot(a,b), 0.0); }
bool isorthogonal(line l1,line l2){ return isorthogonal(vec(l1),vec(l2)); }
//平行
bool isparallel(point a,point b){ return equals(cross(a,b),0.0); }
bool isparallel(line l1,line l2){ return isparallel(vec(l1),vec(l2)); }
//射影
point project(line s,point p){
	point base = vec(s);
	double r=dot(p-s.s,base)/base.norm();
	base = base*r;
	return s.s+base;
}
//反射
point reflect(line l,point p){
	point tmp=project(l,p)-p;
	tmp= tmp*2.0;
	return p+tmp;

}
//交差判定

int ccw(point p0,point p1,point p2){
	point a = p1-p0;
	point b = p2-p0;
if(cross(a,b)>EPS) return 1;//counter_clockwise
if(cross(a,b)<-EPS) return -1;//clockwise
if(dot(a,b)<-EPS)return 2;//online_back
if(a.norm()<b.norm() ) return -2;//online_front
return 0;//on_segment
}
// line p1-p2 line p3-p4
bool intersect(point p1,point p2,point p3,point p4){
	return (ccw(p1,p2,p3)*ccw(p1,p2,p4)<=0 && ccw(p3,p4,p1)*ccw(p3,p4,p2)<=0 ); 
}
// line l1,l2
bool intersect(line l1,line l2){ return intersect(l1.s,l1.e,l2.s,l2.e); }


//距離
//point-point
double distance(point a,point b){ return abs(a-b); }
//point-line(直線）
double distance2(line l,point p){
	return abs(cross(vec(l),p-l.s)/abs(l.e-l.s));
}
//point-line(線分)
double distance(line l,point p){
	if( dot(vec(l),p-l.s) <0.0 ) return abs(p-l.s);
	if( dot(l.s-l.e,p-l.e) <0.0 ) return abs(p-l.e);
	return distance2(l,p);
}
//line-line
double distance(line l1,line l2){
	if(intersect(l1,l2))	return 0.0;
	return min(min(distance(l1,l2.s),distance(l1,l2.e) ), min(distance(l2,l1.s),distance(l2,l1.e) ) );
}

//交点
point crosspoint(line l1,line l2){
	point base = vec(l2);
	double d1 = abs(cross(base,l1.s-l2.s) );
	double d2 = abs(cross(base,l1.e-l2.s) );
	double t = d1/(d1+d2);
	point tmp = vec(l1)*t;
	return l1.s+tmp;
}

int main(){
int q;
cin>>q;
for(int i=0;i<q;i++){
	point p0=make(),p1=make(),p2=make(),p3=make();
	line l0 = make(p0,p1) ,l1 = make(p2,p3);
	printf("%0.10lf\n",distance(l0,l1) );
}

return 0;
}