#include <iostream>
#include <cmath>

#define x first
#define y second

using namespace std;

typedef pair<double, double> PDD;

const int N = 110;
const double eps = 1e-8;

PDD p1[8], p2[8], pos1[N], pos2[N];
double a[11][11];

PDD operator + (PDD a, PDD b)
{
	return { a.x + b.x, a.y + b.y };
}

PDD operator - (PDD a, PDD b)
{
	return { a.x - b.x, a.y - b.y };
}

PDD operator * (PDD a, double b)
{
	return { a.x * b, a.y * b };
}

double operator * (PDD a, PDD b)
{
	return a.x * b.y - a.y * b.x;
}

double operator & (PDD a, PDD b)
{
	return a.x * b.x + a.y * b.y;
}

PDD operator / (PDD a, double b)
{
	return { a.x / b, a.y / b };
}

double get_dist(PDD a, PDD b)
{
	double dx = a.x - b.x, dy = a.y - b.y;
	return sqrt(dx * dx + dy * dy);
}

int gauss(int n)  // 高斯消元，答案存于a[i][n]中，0 <= i < n
{
	int c, r;
	for(c = 0, r = 0; c < n; c ++)
	{
		int t = r;
		for(int i = r; i < n; i ++)  // 找绝对值最大的行
			if(fabs(a[i][c]) > fabs(a[t][c]))
				t = i;

		if(fabs(a[t][c]) < eps) continue;

		for(int i = c; i <= n; i ++) swap(a[t][i], a[r][i]);  // 将绝对值最大的行换到最顶端
		for(int i = n; i >= c; i --) a[r][i] /= a[r][c];  // 将当前行的首位变成1
		for(int i = r + 1; i < n; i ++)  // 用当前行将下面所有的列消成0
			if(fabs(a[i][c]) > eps)
				for(int j = n; j >= c; j --)
					a[i][j] -= a[r][j] * a[i][c];

		r ++;
	}

	if(r < n)
	{
		for(int i = r; i < n; i ++)
			if(fabs(a[i][n]) > eps)
				return 2; // 无解
		return 1; // 有无穷多组解
	}

	for(int i = n - 1; i >= 0; i --)
		for(int j = i + 1; j < n; j ++)
			a[i][n] -= a[i][j] * a[j][n];

	return 0; // 有唯一解
}

void solve()
{
	for(int i = 1; i <= 4; i ++)
		scanf("%lf%lf", &p1[i].x, &p1[i].y);
	for(int i = 1; i <= 4; i ++)
		scanf("%lf%lf", &p2[i].x, &p2[i].y);

	PDD s, t;
	scanf("%lf%lf%lf%lf", &s.x, &s.y, &t.x, &t.y);

	int k, n;
	scanf("%d%d", &k, &n);

	double D = get_dist(p1[1], p1[2]) / get_dist(p2[1], p2[2]);
	// clog << '\n';

	//从 大map 传送到 小map
	auto get = [&](PDD p)
		{
			PDD v = p - p1[1], v1 = p1[2] - p1[1], v2 = p1[4] - p1[1];
			a[0][0] = v1.x, a[0][1] = v2.x, a[0][2] = v.x;
			a[1][0] = v1.y, a[1][1] = v2.y, a[1][2] = v.y;

			// for(int i = 0; i < 2; i ++)
			// {
			// 	for(int j = 0; j <= 2; j ++)
			// 		cout << a[i][j] << " ";
			// 	cout << '\n';
			// }
			// cout << '\n';
			gauss(2);


			double X = a[0][2], Y = a[1][2];
			PDD V1 = p2[2] - p2[1], V2 = p2[4] - p2[1];

			// clog << X << " " << Y << '\n';

			return p2[1] + V1 * X + V2 * Y;
		};

	pos1[0] = s, pos2[0] = t;

	for(int i = 1; i <= n; i ++)
		pos1[i] = get(pos1[i - 1]);
	for(int i = 1; i <= n; i ++)
		pos2[i] = get(pos2[i - 1]);


	// for(int i = 0; i <= n; i ++)
	// {
	// 	clog << pos2[i].x << " " << pos2[i].y << '\n';
	// }
	// clog << '\n';


	double ans = 1e12;
	for(int i = 0; i <= n; i ++)
		for(int j = 0; i + j <= n; j ++)
			ans = min(ans, (i + j) * k + get_dist(pos1[i], pos2[j]));

	printf("%.10lf\n", ans);
}

int main()
{
	int T;
	scanf("%d", &T);

	while(T --)
		solve();

	return 0;
}