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<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<iostream>
#include<set>
#include<map>
#define re register
#define ll long long
using namespace std;
inline int gi(){
	int f=1,sum=0;char ch=getchar();
	while(ch>'9'|| ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
	return sum*f;
}
const double eps=1e-10,Pi=acos(-1.0);
struct node{
	double x,y;
	node operator+(node b){return (node){x+b.x,y+b.y};}
	node operator-(node b){return (node){x-b.x,y-b.y};}
	node operator*(double b){return (node){x*b,y*b};}
	double len(){return sqrt(x*x+y*y);}
};
int fh(double a){return fabs(a)<eps?0:a>eps?1:-1;}
double cross(node a,node b){return a.x*b.y-a.y*b.x;}
double dot(node a,node b){return a.x*b.x+a.y*b.y;}
node projection(node p1,node p2,node p){node v=p2-p1;return p1+v*(dot(v,p-p1)/dot(v,v));}
node reflection(node p1,node p2,node p){node P=projection(p1,p2,p);P=P*2-p;return P;}
void Counter_Clockwise(node p1,node p2,node p){
	if(cross(p2-p1,p-p1)>eps){puts("COUNTER_CLOCKWISE");return;}
	if(cross(p2-p1,p-p1)<-eps){puts("CLOCKWISE");return;}
	if(dot(p2-p1,p-p1)<-eps){puts("ONLINE_BACK");return;}
	if(dot(p2-p1,p2-p1)>=dot(p-p1,p2-p1)){puts("ON_SEGMENT");return;}
	puts("ONLINE_FRONT");
}
bool insegment(node A,node B,node p){
	if(fh(cross(B-A,p-A)))return false;
	return ((fh(p.x-A.x)>=0 && fh(B.x-p.x)>=0) || (fh(p.x-A.x)<=0 && fh(B.x-p.x)<=0)) && ((fh(p.y-A.y)>=0 && fh(B.y-p.y)>=0) || (fh(p.y-A.y)<=0 && fh(B.y-p.y)<=0));
}
int parallel(node A1,node A2,node B1,node B2){
	if(!fh(cross(B2-B1,A2-A1)))return 2;
	if(!fh(dot(B2-B1,A2-A1)))return 1;
	return 0;
}
bool intersection(node A1,node A2,node B1,node B2){
	if(!fh(cross(B2-B1,A2-A1))){
		if(insegment(A1,A2,B1) || insegment(A1,A2,B2) || insegment(B1,B2,A1) || insegment(B1,B2,A2))return true;
		return false;
	}
	return fh(cross(A1-B1,B2-B1)*cross(A2-B1,B2-B1))<=0 && (fh(cross(B1-A1,A2-A1)*cross(B2-A1,A2-A1))<=0);
}
node getintersection(node A1,node A2,node B1,node B2){
	node v=A2-A1;
	return A1+v*(cross(B2-B1,B1-A1)/cross(B2-B1,A2-A1));
}
double distance(node A1,node A2,node B1,node B2){
	if(intersection(A1,A2,B1,B2))return 0.0;
	node p;double ans=1e18;
	ans=min(ans,(A1-B1).len());ans=min(ans,(A1-B2).len());ans=min(ans,(A2-B1).len());ans=min(ans,(A2-B2).len());
	p=projection(A1,A2,B1);
	if(insegment(A1,A2,p))ans=min(ans,(B1-p).len());
	p=projection(A1,A2,B2);
	if(insegment(A1,A2,p))ans=min(ans,(B2-p).len());
	p=projection(B1,B2,A1);
	if(insegment(B1,B2,p))ans=min(ans,(A1-p).len());
	p=projection(B1,B2,A2);
	if(insegment(B1,B2,p))ans=min(ans,(A2-p).len());
	return ans;
}
int main(){
	node p1,p2,p3,p4;
	int Q=gi();
	while(Q--){
		p1.x=gi();p1.y=gi();p2.x=gi();p2.y=gi();
		p3.x=gi();p3.y=gi();p4.x=gi();p4.y=gi();
		printf("%.8lf\n",distance(p1,p2,p3,p4));
	}
	return 0;
}