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;
typedef complex<double> Point;
typedef pair<Point, Point> Line;
const double EPS = 1e-10;

// 内積 |a||b|cosθ
double dot(Point a, Point b){
	// return a.real() * b.real() + a.imag() * b.imag();
	return (conj(a) * b).real();
}
// 外積、符号付面積の２倍　|a||b|sinθ
double cross(Point a, Point b){
	// return a.real() * b.imag() - a.imag() * b.real();
	return (conj(a) * b).imag();
}
// 点の位置関係　2点a,bから見た点cの位置関係
int ccw(Point a, Point b, Point c){
    // COUNTER_CLOCKWISE
    if(cross(b - a, c - a) > EPS) return 1;
    // CLOCKWISE
    if(cross(b - a, c - a) < -EPS) return -1;
    // ONLINE_BACK
    if(dot(b - a, c - a) < -EPS) return 2;
    // ONLINE_FRONT
    if(abs(b - a) + EPS < abs(c - a)) return -2;
    // ON_SEGMENT
    return 0;
}
// 点と直線の距離
double line_to_point_dis(Line l, Point p){
    Point a = l.first, b = l.second, c = p;
    return fabs(cross(c - a, b - a)) / fabs(abs(b - a));
}
// 点と線分の距離
double seg_to_point_dis(Line l, Point p){
    Point a = l.first, b = l.second, c = p;
    if(dot(b - a, c - a) < EPS) return abs(c - a);
    if(dot(a - b, c - b) < EPS) return abs(c - b);
    return line_to_point_dis(l, p);
}
// 線分の交差判定
bool is_cross(Line a, Line b){
    if(ccw(a.first, a.second, b.first) * ccw(a.first, a.second, b.second) <= 0 && ccw(b.first, b.second, a.first) * ccw(b.first, b.second, a.second) <= 0){
        return true;
    }
    return false;
}
// 線分と線分の距離
double seg_to_seg_dis(Line a, Line b){
    double res = seg_to_point_dis(a, b.first);
    res = min(res, seg_to_point_dis(a, b.second));
    res = min(res, seg_to_point_dis(b, a.first));
    res = min(res, seg_to_point_dis(b, a.second));
    // 線分aと線分bが交差している場合
    if(is_cross(a, b)) res = 0.0;
    return res;
}

int main(){
    int q;
    cin >> q;
    for(int i = 0; i < q; i++){
        double x, y;
        Point p[4];
        for(int j = 0; j < 4; j++){
            cin >> x >> y;
            p[j] = {x, y};
        }
        Line s1 = {p[0], p[1]};
        Line s2 = {p[2], p[3]};
        double len = seg_to_seg_dis(s1, s2);
        printf("%.10f\n", len);
    }
}