#include<cmath>

#include<cstdio>

#include<cstring>

#include<algorithm>

#define max(a,b) (a) > (b) ? (a) : (b)

#define N 1003

using namespace std;

const double Pi = acos(-1);

inline int dcmp(double x) {return (fabs(x) < 1e-6) ? 0 : (x < 0 ? -1 : 1);}

inline double sqr(double x) {return x * x;}

struct Point {

	double x, y;

	Point(double _x = 0, double _y = 0) : x(_x), y(_y) {}

};

struct Rround {

	Point O; double R;

	Rround(Point _O = Point(0, 0), double _R = 0) : O(_O), R(_R) {}

};

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

inline double dis(Rround a, Rround b) {return sqrt(sqr(a.O.x - b.O.x) + sqr(a.O.y - b.O.y));}

inline bool cover(Rround to, Rround now) {return to.R >= now.R + dis(to, now);}

struct Splay {

	double l, r;

	Splay(double _l = 0, double _r = 0) : l(_l), r(_r) {}

};

Splay s[N];

int n, top;

double ans = 0, tt, nxt;

Rround r[N];

bool pd;

inline void ins(Rround a, Rround b) {

	double d = dis(a, b), t, st, l;

	t = (sqr(a.R) - sqr(b.R) + sqr(d)) / (d + d);

	st = atan2((a.O.y - b.O.y), (a.O.x - b.O.x));

	//st = atan2((a.O.x - b.O.x), (a.O.y - b.O.y));

	l = acos(t / a.R);

	s[++top] = Splay(st - l, st + l);

}

inline bool cmp(Splay a, Splay b) {return a.l < b.l;}

int main() {

	scanf("%d", &n);

	for(int i = 1; i <= n; ++i)

		scanf("%lf%lf%lf", &r[i].R, &r[i].O.x, &r[i].O.y);

	for(int i = 1; i <= n; ++i) {

		pd = 0;

		for(int j = i + 1; j <= n; ++j)

			if (cover(r[j], r[i])) {

				pd = 1;

				break;

			}

		if (pd)

			continue;

		top = 0;

		for(int j = i + 1; j <= n; ++j)

			if (!cover(r[i], r[j]) && r[i].R + r[j].R >= dis(r[i], r[j]))

				ins(r[i], r[j]);

		for(int j = 1; j <= top; ++j) {

			if (s[j].l < 0) s[j].l += 2 * Pi;

			if (s[j].r < 0) s[j].r += 2 * Pi;

			if (s[j].l > s[j].r) {

				s[++top] = Splay(0, s[j].r);

				s[j].r = 2 * Pi;

			}

		}

		sort(s + 1, s + top + 1, cmp);

		tt = 0; nxt = 0;

		for(int j = 1; j <= top; ++j)

			if (s[j].l > nxt) {

				tt += s[j].l - nxt;

				nxt = s[j].r;

			} else

				nxt = max(nxt, s[j].r);

		tt += 2 * Pi - nxt;

		ans += tt * r[i].R;

	}

	printf("%.3lf\n",ans);

	return 0;

}