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 <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#define EPS (1e-10)
#define equals(a,b) (fabs(a)-fabs(b)<EPS)


using namespace std;

class Point{
public:
    double x,y;

    Point(double x=0,double y=0):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 * (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;
    }
};

typedef Point Vector;
///Program 16.3 表示线段的结构体
struct Segment{
    Point p1,p2;
};

///Program 16.4 表示直线的结构体
typedef Segment Line;

class Circle{
public:
    Point c;
    double r;
    Circle(Point c=Point(),double r=0.0):c(c),r(r){}
};

///Program 16.6 多边形
typedef vector<Point> Polygon;

///内积
double dot(Vector a,Vector b){
    return a.x*b.x+a.y*b.y;
}
///外积
double cross(Vector a,Vector b){
    return a.x*b.y-a.y*b.x;
}

double norm(Point p) { return p.x*p.x+p.y*p.y; }
double abs(Point p) { return sqrt(norm(p)); }
//double norm() { return x*x+y*y; }
int ccw(Point p0,Point p1,Point p2){
    Vector a=p1-p0;
    Vector b=p2-p0;
    if(cross(a,b)>EPS)  {//printf("COUNTER_CLOCKWISE\n");
            return 1;}///p2位于p0-->p1的逆时针方向
    if(cross(a,b)<-EPS) {//printf("CLOCKWISE\n");
        return -1;}///p2位于p0-->p1的顺时针方向
    if(dot(a,b)<-EPS)   {//printf("ONLINE_BACK\n");
        return 2;}///p2位于线段p0p1后方
    if(a.norm()<b.norm())  { //printf("ONLINE_FRONT\n");
        return -2;}///p2位于线段p0p1前方

  //  printf("ON_SEGMENT\n");///p2位于线段p0p1的中间
    return 0;
}

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);
}

bool intersect(Segment s1,Segment s2){
    return intersect(s1.p1,s1.p2,s2.p1,s2.p2);
}


///两点之间的距离
double getDistance(Point a,Point b){
    return abs(a-b);
}

///点与直线的距离
double getDistanceLP(Line l,Point p){
    return abs(cross(l.p2-l.p1,p-l.p1)/abs(l.p2-l.p1));
}

///点到线段的距离
double getDistanceSP(Segment s,Point p){
    if(dot(s.p2-s.p1,p-s.p1)<0.0)   return abs(p-s.p1);
    if(dot(s.p1-s.p2,p-s.p2)<0.0)   return abs(p-s.p2);
    return getDistanceLP(s,p);
}

///线段到线段的距离
double getDistance(Segment s1,Segment s2){
    if(intersect(s1,s2))    return 0.0;
    return min(min(getDistanceSP(s1,s2.p1),getDistanceSP(s1,s2.p2)),
               min(getDistanceSP(s2,s1.p1),getDistanceSP(s2,s1.p2)));
}


int main(){
    Segment s1,s2;
    int q;
    scanf("%d",&q);
    for(int i=0;i<q;i++){
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&s1.p1.x,&s1.p1.y,&s1.p2.x,&s1.p2.y,&s2.p1.x,&s2.p1.y,&s2.p2.x,&s2.p2.y);
        printf("%.10f\n",getDistance(s1,s2));
    }
    return 0;
}