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;
const double eps=1e-9;
struct dot{
	double x,y;
	double len(){return sqrt(x*x+y*y);}
	dot(double a=0.0,double b=0.0){x=a,y=b;}
	dot operator +(const dot&b)const {return dot(x+b.x,y+b.y);}
	dot operator -(const dot&b)const {return dot(x-b.x,y-b.y);}
	dot operator *(const double&z)const {return dot(x*z,y*z);}
	bool operator <(const dot&b)const {return (x<b.x)||(x==b.x&&y<b.y);}
	void read(){scanf("%lf%lf",&x,&y);} 
};
struct line{
	dot x1,x2;
	void read(){x1.read();x2.read();};
	dot getdot(){return x2-x1;}
};
double chaji(dot x,dot y){
	return x.x*y.y-x.y*y.x;
}
double neiji(dot x,dot y){
	return x.x*y.x+x.y*y.y;
}
double cos(dot x,dot y){
	if (!x.len()||!y.len())return 0;
	return neiji(x,y)/x.len()/y.len();
}
dot Projection(dot x,line y){//求x关于直线y的投影点（垂足） 
	y.x2=y.x2-y.x1;
	x=x-y.x1;
	double len=cos(x,y.x2)*x.len()/y.x2.len();
	y.x2=y.x2*len;
	return y.x2+y.x1;
}
dot Reflection(dot x,line y){//求x关于直线y的对称点 
	dot z=Projection(x,y);
	return z*2-x;
}
int Counter_Clockwise(dot x,line y){//求两个向量的五种关系 
	x=x-y.x1;y.x2=y.x2-y.x1;
	if (chaji(y.x2,x)!=0){
		if (chaji(y.x2,x)>0)return 1;
		return 2;
	} 
	if (fabs(cos(x,y.x2)+1)<eps)return 3;
	if (x.len()>y.x2.len())return 4;
	return 5;
} 
int Parallel_Orthogonal(line x,line y){//求两个直线是垂直还是平行还是其他 
	dot a=x.getdot(),b=y.getdot();
	if (chaji(a,b)==0)return 2;
	if (neiji(a,b)==0)return 1;
	return 0;
}
int Intersection(line x,line y){//求两条直线是否相交 
	int Minx1=min(x.x1.x,x.x2.x),Maxx1=max(x.x1.x,x.x2.x);
	int Minx2=min(y.x1.x,y.x2.x),Maxx2=max(y.x1.x,y.x2.x);
	int Miny1=min(x.x1.y,x.x2.y),Maxy1=max(x.x1.y,x.x2.y);
	int Miny2=min(y.x1.y,y.x2.y),Maxy2=max(y.x1.y,y.x2.y);
	if (Minx1>Maxx2||Minx2>Maxx1)return 0;
	if (Miny1>Maxy2||Miny2>Maxy1)return 0;
	if (chaji(x.getdot(),y.x1-x.x1)*chaji(x.getdot(),y.x2-x.x1)>0)return 0;
	if (chaji(y.getdot(),x.x1-y.x1)*chaji(y.getdot(),x.x2-y.x1)>0)return 0;
	return 1;
}
dot CrossPoint(line x,line y){//求两条线段的交点 
	if (fabs(cos(x.getdot(),y.getdot())-1)<eps){
		if (x.x2<y.x2)return y.x2;
		else return x.x2;
	}
	else {
		if (x.x2.x-x.x1.x==0){
			double k2=(y.x2.y-y.x1.y)/(y.x2.x-y.x1.x),b2=y.x2.y-y.x2.x*k2;
			return dot(x.x2.x,x.x2.x*k2+b2);
		}
		if (y.x2.x-y.x1.x==0){
			double k1=(x.x2.y-x.x1.y)/(x.x2.x-x.x1.x),b1=x.x2.y-x.x2.x*k1;
			return dot(y.x2.x,y.x2.x*k1+b1);			
		}
		double k1=(x.x2.y-x.x1.y)/(x.x2.x-x.x1.x),b1=x.x2.y-x.x2.x*k1;
		double k2=(y.x2.y-y.x1.y)/(y.x2.x-y.x1.x),b2=y.x2.y-y.x2.x*k2;
		dot ans;
		ans.x=(b2-b1)/(k1-k2);ans.y=ans.x*k1+b1;
		return ans;
	}
} 
double Dis(dot x,dot y){//点到点的距离 
	return sqrt((x.x-y.x)*(x.x-y.x)+(x.y-y.y)*(x.y-y.y));
}
double dist(line x,dot y){//点到直线的距离 
	dot P=Projection(y,x);
	double k=neiji(x.getdot(),P-x.x1)/neiji(x.getdot(),x.getdot());
	if (k<=0)return Dis(x.x1,y);
	if (k>=1)return Dis(x.x2,y);
	return Dis(y,P);
}
double Distance(line x,line y){
	return min(min(dist(x,y.x1),dist(x,y.x2)),min(dist(y,x.x1),dist(y,x.x2)));
}
int main(){
	int T;
	scanf("%d",&T);
	while (T--){
		line x,y;
		x.read();y.read();
		if (Intersection(x,y))puts("0.0000000000");
		else printf("%.10lf\n",Distance(x,y));
	}
	return 0;
}