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 <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
using namespace std;
int num;
double x,yy1,yy2,yy3,yy4;
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
int sgn(double x){
    if(fabs(x)<eps){
        return 0;
    }
    if(x<0){
        return -1;
    }else{
        return 1;
    }
}
struct Point{
    double x,y;
    Point(){}
    void input(){
        scanf("%lf%lf",&x,&y);
    }
    Point(double _x,double _y){
        x = _x;y = _y;
    }
    bool operator < (Point b)const{
        return sgn(x-b.x)==0?sgn(y-b.y)<0:x<b.x;
    }
    Point operator - (const Point &b)const{
        return Point(x-b.x,y-b.y);
    }
    double operator *(const Point &b)const{
        return x*b.x+y*b.y;
    }
    double operator ^(const Point &b)const{
        return x*b.y-y*b.x;
    }
    double distance(Point p){
        return hypot(x-p.x,y-p.y);
    }
};
struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){
        s = _s;e = _e;
    }
    // 线段长度
    double length(){
        return s.distance(e);
    }
    // 判断线段是否相交
    int segcrossseg(Line v){
        int d1 = sgn((e-s)^(v.s-s));
        int d2 = sgn((e-s)^(v.e-s));
        int d3 = sgn((v.e-v.s)^(s-v.s));
        int d4 = sgn((v.e-v.s)^(e-v.s));
        if((d1^d2) == -2 && (d3^d4)==-2){
            return 2;
        }
        return (d1==0&&sgn((v.s-s)*(v.s-e))<=0)||(d2==0&&sgn((v.e-s)*(v.e-e))<=0)||(d3==0&&sgn((s-v.s)*(s-v.e))<=0)||(d4==0&&sgn((e-v.s)*(e-v.e))<=0);
    }
    // 点线距离
    double dispointtoline(Point p){
        return fabs((p-s)^(e-s))/length();//面积除以高
    }
    // 点到线段距离（当是钝角的时候，距离不是延长线，而是到两个端点的最小值）
    double dispointtoseg(Point p){
        if(sgn((p-s)*(e-s))<0||sgn((p-e)*(s-e))<0){
            return min(p.distance(s),p.distance(e));
        }
        return dispointtoline(p);
    }
    // 线段到线段的距离
    // 用之前判断是否相交
    double dissegtoseg(Line v){
        return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
    }
};
int main(){
    int q;
    scanf("%d",&q);
    Point p0,p1,p2,p3;
    while(q--){
        p0.input();
        p1.input();
        p2.input();
        p3.input();
        Line line1(p0,p1);
        Line line2(p2,p3);
        double ans = 0;
        if(line1.segcrossseg(line2) == 0){
            ans = line1.dissegtoseg(line2);
        }
        printf("%.10f\n",ans);
    }

    return 0;
}