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 long long ll;
typedef pair<ll, ll> lpair;
const ll MOD = 1e9 + 7;
const ll INF = 1e18;
#define rep(i,m,n) for(ll i = (m); i < (n); i++)
#define rrep(i,m,n) for(ll i = (m); i >= (n); i--)
#define print(x) cout << (x) << endl;
#define print2(x,y) cout << (x) << " " << (y) << endl;
#define printa(x,n) for(ll i = 0; i < n; i++){ cout << (x[i]) << " ";} cout<<endl;

#define EPS (1e-10)
#define EQ(a,b) (abs((a) - (b)) < EPS)
typedef complex<double> P;
double dot(P a, P b){ //内積
    return (a.real() * b.real() + a.imag() * b.imag());
}

double cross(P a, P b){ //外積
    return (a.real() * b.imag() - a.imag() * b.real());
}

bool is_orth(P a1, P a2, P b1, P b2){ //垂直か
    return EQ(dot(a1 - a2, b1 - b2), 0.0);
}

bool is_para(P a1, P a2, P b1, P b2){ //平行か
    return EQ(cross(a1 - a2, b1 - b2), 0.0);
}

bool is_online(P a, P b, P c){ //cが線分上か
    return (abs(a-c) + abs(c-b) < abs(a-b) + EPS);
}

bool distance_lp(P a, P b, P c){ //直線abと点cの距離
    return abs(cross(b-a, c-a)) / abs(b-a);
}

double norm(P a){
  return a.real() * a.real() +a.imag() *a.imag();
}

ll ccw(P p0,P p1,P p2){
  P a = p1-p0;
  P b = p2-p0;
  if(cross(a,b) > EPS) return 1;
  if(cross(a,b) < -EPS) return -1;
  if(dot(a,b) < -EPS) return 2;
  if(abs(a) < abs(b)) return -2;
  return 0;
}

bool is_intersect(P p1, P p2, P p3, P p4){
  return (ccw(p1,p2,p3) * ccw(p1,p2,p4) <= 0 &&
      ccw(p3,p4,p1) * ccw(p3,p4,p2) <= 0 );
}

P intersect_pos(P a1, P a2, P b1, P b2){ //直線同士の交点
    P a = a2 - a1;
    P b = b2 - b1;
    return a1 + a * cross(b, b1 - a1) / cross(b, a);
}

double getDistanceLP(P a1, P a2, P b){
    return abs(cross(a2-a1, b-a1) / abs(a2 - a1));
}

double getDistanceSP(P a1, P a2, P b){
    if(dot(a2 - a1, b - a1) < EPS) return abs(b - a1);
    if(dot(a1 - a2, b - a2) < EPS) return abs(b - a2);
    return getDistanceLP(a1, a2, b);
}
double getDistance(P a1, P a2, P b1, P b2){
    if(is_intersect(a1, a2, b1, b2)) return 0.0;
    double v1 = min(getDistanceSP(a1, a2, b1), getDistanceSP(a1, a2, b2));
    double v2 = min(getDistanceSP(b1, b2, a1), getDistanceSP(b1, b2, a2));
    return min(v1,v2);
}

#define printP(P) cout << "(" << P.real() << ", " << P.imag() << ")" << endl;




int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll q;
    cin >> q;
    while(q--){
        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);
        double ans = getDistance(p0, p1, p2, p3);
        cout << setprecision(16) << ans << endl;
    }
}