/*
 * Warmup Contest #2, September 4, 2010 Problem F 
 * Solution by Nathaniel Barshay
 * Originally from http://2008.nwerc.eu/contest/problemset (Problem I) 
 * ACM NWERC 2008 Problem I
 * 
 * The distance between two polygons is the closest distance between
 * any pair of lines one of each polygon. (You can visualize this and
 * prove it to yourself.)
 *
 * Nathaniel has good geometry routines here to make his life easy. 
 *
 */


#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <deque>
#include <bitset>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cctype>
#include <queue>
#include <complex>
#include <queue>
#include <string.h>
#include <fstream>
#include <stack>
#include <assert.h>
#include <time.h>
using namespace std;

#define IT(c) typeof((c).begin())

#define For(i, a, b) for(int (i) =  int(a); i < int(b); ++i)
#define rep(x, n) For(x,0,n)
#define forit(c, i) for(IT(c) i = (c).begin(); i != (c).end(); ++i)

#define bound(num, lower, upper) (max(min((num),((upper)-1)),(lower)))
#define debug(x) cerr << #x << " = " << x << "\n"

#define sz(a) int((a).size())
#define pb push_back
#define all(c) (c).begin(),(c).end()
#define mp make_pair

#define setmin(a,b) a=min(a,b)
#define setmax(a,b) a=max(a,b)

typedef vector<int> vi;
typedef vector<vector<int> > vvi;
typedef vector<string> vs;
typedef pair<int, int> ii;

/*==================================================================================================================*/

int dot(int a[], int b[], int c[])
{
	int ab[2];
	int bc[2];
	ab[0] = b[0] - a[0];
	ab[1] = b[1] - a[1];
	bc[0] = c[0] - b[0];
	bc[1] = c[1] - b[1];
	return (ab[0]*bc[0])+ (ab[1]*bc[1]);
}

int cross(int a[], int b[], int c[])
{
	int ab[2];
	int ac[2] ;
	ab[0] = b[0] - a[0];
	ab[1] = b[1] - a[1];
	ac[0] = c[0] - a[0];
	ac[1] = c[1] - a[1];
	return (ab[0]*ac[1]) - (ab[1]*ac[0]);
}

double dist(int a[], int b[])
{
	int da=a[0]-b[0];
	int db=a[1]-b[1];
	return sqrt(da*da+db*db);
}

double ldist(int a[], int b[], int c[])
{
	double raw=cross(a,b,c)/dist(a,b);

	int dott=dot(a,b,c);
	if(dott>0) return dist(b,c);
	dott=dot(b,a,c);
	if(dott>0) return dist(a,c);
	return abs(raw);
}

int main()
{
int np;cin>>np;
rep(tp,np)
{
	int n; cin>>n;
	vector<int> ax;
	vector<int> ay;
	rep(i,n)
	{
		int t;cin>>t;
		ax.pb(t);
		cin>>t;
		ay.pb(t);
	}
	int m;cin>>m;
	vector<int> bx;
	vector<int> by;
	rep(i,m)
	{
		int t;cin>>t;
		bx.pb(t);
		cin>>t;
		by.pb(t);
	}
	double res=1000000;
	rep(tt,2)
	{
		vector<int>& px=tt?ax:bx;
		vector<int>& py=tt?ay:by;
		vector<int>& lx=tt?bx:ax;
		vector<int>& ly=tt?by:ay;
		rep(i,sz(px))
		{
			int p[2];
			p[0]=px[i]; p[1]=py[i];
			rep(k,sz(lx))
			{
				int a[2];
				a[0]=lx[k]; a[1]=ly[k];
				int b[2];
				b[0]=lx[(k+1)%sz(lx)]; b[1]=ly[(k+1)%sz(ly)];
				setmin(res, ldist(a,b,p));
			}
		}
	}
	cout<<res/2<<endl;
}
}