#include <stdio.h>
#include <string.h>

#include <vector>
#include <algorithm>

#define LOG(FMT...) fprintf(stderr, FMT)
#define eb emplace_back

using namespace std;

typedef long double LDB;
typedef long long LL;

int zA, zB;
template <typename T>
struct Vec {
	T x, y;
	Vec operator - (const Vec &t) const {
		return (Vec){x - t.x, y - t.y};
	}
	T operator & (const Vec &t) const {
		return x * t.x + y * t.y;
	}
	T operator * (const Vec &t) const {
		return x * t.y - y * t.x;
	}
	Vec<LDB> operator * (LDB t) const {
		return (Vec<LDB>){x * t, y * t};
	}
	bool operator == (const Vec &t) const {
		return x == t.x && y == t.y;
	}
	Vec rot() const {
		return (Vec){-y, x};
	}
	T md2() const {
		return x * x + y * y;
	}
};

Vec<int> A, B, p, q;

int solve() {
	if (A == B) {
		return 2;
	}
	Vec<int> AB = B - A;
	if (!(p * q)) {
		return (AB & p) ? 4 : 2;
	}
	if (zA != zB && ((p & AB) == 0 || (q & AB) == 0)) {
		return 3;
	}
	if (zA == zB && (!(AB & p) || !(AB & q))) {
		return 4;
	}
	Vec<int> rp = p.rot(), rq = q.rot();
	Vec<LDB> C = rp * (AB * rq / (LDB)(rp * rq));
	C.x += A.x, C.y += A.y;
	LDB s = (C.x - A.x) * (C.x - B.x) + (C.y - A.y) * (C.y - B.y);
	return zA * zA >= 4 * s ? 3 : 4;
}

int main() {
	int task;
	for (scanf("%d", &task); task--; ) {
		scanf("%d%d%d%d%d%d%d%d%d%d", &A.x, &A.y, &zA, &p.x, &p.y, &B.x, &B.y, &zB, &q.x, &q.y);
		zA -= zB, zB = 0;
		printf("%d\n", solve());
	}
	return 0;
}