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)

inline int sign(const double &x){
    if(x>EPS) return 1;
    else if(x<-EPS) return -1;
    return 0;
}

/*   点  */
struct Point{
    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 k) { return Point(k*x,k*y); }
    Point operator / (double k) { return Point(x/k,y/k); }

    //向量的大小
    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;
    }
};

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

/*   向量  */
typedef Point Vector;

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

/*   线段  */
struct Segment{
    Point p1,p2;
};

/*   直线  */
typedef Segment Line;

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

//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 (p-s.p1).abs();
    if(dot(s.p1-s.p2,p-s.p2)<0.0) return (p-s.p2).abs();
    return getDistanceLP(s,p);
}

//判断p2与向量p1-p0的位置关系

const int COUNTER_CLOCKWISE = -1;   //逆时针
const int CLOCKWISE = 1;            //顺时针
const int ON_SEGMENT = 0;           //在线段上

int ccw(Point p0,Point p1,Point p2){
    Vector a=p1-p0;
    Vector b=p2-p0;
    if(cross(a,b)>EPS) return COUNTER_CLOCKWISE;
    if(cross(a,b)<-EPS) return CLOCKWISE;
    if(dot(a,b)<-EPS) return 2;
    if(a.norm()<b.norm()) return -2;
    return ON_SEGMENT;
}

//判断线段p1p2与线段p3p4是否相交
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);
}

//线段与线段的距离
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)));
}

/* 直线正交
 * 判断向量a，b是否正交 =》a，b的内积为0
 */
bool isOrthogonal(Vector a,Vector b){
    return equals(dot(a,b),0.0);
}

bool isOrthogonal(Point a1,Point a2,Point b1,Point b2){
    return isOrthogonal(a1-a2,b1-b2);
}

/* 直线平行
 * 判断向量a，b是否平行 =》a，b的内积为0
 */
bool isParallel(Vector a,Vector b){
    return equals(cross(a,b),0.0);
}

bool isParallel(Point a1,Point a2,Point b1,Point b2){
    return isParallel(a1-a2,b1-b2);
}

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

/*   多边形  */
typedef vector<Point> Polygon;

int main(){
    int n;
    scanf("%d",&n);
    while(n--){
        Point p1,p2,p3,p4;
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y,&p3.x,&p3.y,&p4.x,&p4.y);
        printf("%.10f\n",getDistance(Segment{p1,p2},Segment{p3,p4}));
    }
    return 0;
}