/*
对于每个圆, 考虑之后所有的圆覆盖了他的哪些部分即可
所以我们要实现两圆求交点, 区间求并
*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#include<cmath>
#define ll long long
#define M 2010
#define mmp make_pair
using namespace std;
const double pi = acos(-1.0);
int read() {
int nm = 0, f = 1;
char c = getchar();
for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
return nm * f;
}
struct C {
double x, y, r;
} cir[M];
struct Note {
double p;
int v;
bool operator < (const Note &b) const {
return this->p < b.p;
}
} note[M];
int n;
double ans, diss;
double dis(C a, C b) {
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
int main() {
n = read();
for(int i = 1; i <= n; i++) scanf("%lf%lf%lf", &cir[i].r, &cir[i].x, &cir[i].y);
for(int i = 1; i <= n; i++) {
int f = false, tp = 0, sum = 0;
for(int j = i + 1; j <= n; j++) {
diss = dis(cir[i], cir[j]);
if(diss + cir[i].r <= cir[j].r) {
f = true;
break;
}
if(diss > cir[i].r + cir[j].r) continue;
if(diss + cir[j].r <= cir[i].r) continue;
// if(diss <= fabs(cir[i].r - cir[j].r)) continue;
double a = acos((cir[i].r * cir[i].r + diss * diss - cir[j].r * cir[j].r) / (2.0 * cir[i].r * diss));
double b = atan2(cir[j].y - cir[i].y, cir[j].x - cir[i].x);
note[++tp] = (Note) {
b - a, 1
};
note[++tp] = (Note) {
b + a, -1
};
if(note[tp - 1].p < 0) note[tp - 1].p += 2 * pi;
if(note[tp - 1].p > 2 * pi) note[tp - 1].p -= 2 * pi;
if(note[tp].p < 0) note[tp].p += 2 * pi;
if(note[tp].p > 2 * pi) note[tp].p -= 2 * pi;
if(note[tp - 1].p > note[tp].p) sum++;
}
if(f) continue;
ans += cir[i].r * 2 * pi;
if(!tp) continue;
sort(note + 1, note + tp + 1);
for(int j = 1; j <= tp; j++) {
if(sum) ans -= cir[i].r * (note[j].p - note[j - 1].p);
sum += note[j].v;
}
if(sum) ans -= cir[i].r * (2 * pi - note[tp].p);
}
printf("%.3lf
", ans);
return 0;
}
/*
5
0.743 0.247 -0.594
0.051 0.837 0.239
0.257 0.617 -0.676
0.780 -0.119 -0.547
0.091 0.318 -0.661
*/