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<complex>
#include<iostream>
#include<algorithm>

using namespace std;

typedef complex<long double> P;

// ??§????????°INF
#define INF (1e16)
// ?¨±?????????????????
#define EPS (1e-16)
// 2??????????????????????????????????????????
#define EQ(a,b) (abs((a)-(b)) < EPS)
// 2??????????????????????????????????????????
#define EQV(a,b) ( EQ((a).real(), (b).real()) && EQ((a).imag(), (b).imag()) )

// ?????? (dot product) : a???b = |a||b|cos??
long double dot(P a, P b) {
	return (a.real() * b.real() + a.imag() * b.imag());
}

// ?????? (cross product) : a??b = |a||b|sin??
long double cross(P a, P b) {
	return (a.real() * b.imag() - a.imag() * b.real());
}

// ?????????????????????
int is_intersected_ls(P a1, P a2, P b1, P b2) {
	return (cross(a2 - a1, b1 - a1) * cross(a2 - a1, b2 - a1) < EPS) &&
		(cross(b2 - b1, a1 - b1) * cross(b2 - b1, a2 - b1) < EPS);
}

// 2??´?????????????????? : a//b <=> cross(a, b) = 0
int is_parallel(P a1, P a2, P b1, P b2) {
	return EQ(cross(a1 - a2, b1 - b2), 0.0);
}

// ???c?????????a,b????????????????????????(1)
int is_point_on_line(P a, P b, P c) {
	return EQ(cross(b - a, c - a), 0.0) &&
		(dot(b - a, c - a) > -EPS) &&
		(dot(a - b, c - b) > -EPS);
}

// a1,a2???????????¨??????????????¨b1,b2???????????¨??????????????????????¨????
P intersection_ls(P a1, P a2, P b1, P b2) {
	P b = b2 - b1;
	long double d1 = abs(cross(b, a1 - b1));
	long double d2 = abs(cross(b, a2 - b1));
	long double t = d1 / (d1 + d2);

	return a1 + (a2 - a1) * t;
}


// ???a,b???????????¨??????????????¨???c??¨????????¢
long double distance_ls_p(P a, P b, P c) {
	if (dot(b - a, c - a) < EPS) return abs(c - a);
	if (dot(a - b, c - b) < EPS) return abs(c - b);
	return abs(cross(b - a, c - a)) / abs(b - a);
}

int main() {
	int q; cin >> q;
	for (int i = 0; i < q; i++) {
		long double x0, y0, x1, y1, x2, y2, x3, y3;
		cin >> x0 >> y0 >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
		P p0(x0, y0), p1(x1, y1), p2(x2, y2), p3(x3, y3);
		if (is_parallel(p0, p1, p2, p3)) {
			long double d0 = min(distance_ls_p(p0, p1, p2), distance_ls_p(p0, p1, p3));
			long double d1 = min(distance_ls_p(p2, p3, p0), distance_ls_p(p2, p3, p1));
			long double ans = min(d0, d1);

			cout.precision(20);
			cout << showpoint << ans << endl;
			continue;
		}
		if (is_intersected_ls(p0, p1, p2, p3)) {
			cout.precision(20);
			cout << showpoint << 0.00000000000 << endl;
			continue;
		}
		long double d0 = min(distance_ls_p(p0, p1, p2), distance_ls_p(p0, p1, p3));
		long double d1 = min(distance_ls_p(p2, p3, p0), distance_ls_p(p2, p3, p1));
		long double ans =min(d0,d1);

		cout.precision(20);
		cout << showpoint << ans << endl;
	}
}