Problem: A crop circle suddenly appeared on the vast agricultural land of Argentina. A total of n crop circles were confirmed, including overlapping, popping, large, and small ones.

When a mystery hunter tried to capture the whole picture of a crop circle from the air, he found it difficult to show the beautiful pattern in the image because the outline of each circle was not clear.

Therefore, the mystery hunter proposed to emphasize the contour by installing a string of special material along the contour.

The part where the string is installed is the part on the circumference of one crop circle that is not included in any other crop circle.

The film crew asked you to create a program to measure the required string length. Enter the center coordinates and radius of each crop circle and create a program to report the length of the string to be installed.

For reference, the first and second cases of the input / output examples are shown in FIGS. 1 and 2, respectively. The part of the string is shown by a thick line.


<image>


Figure 1



<image>


Figure 2



Constraints

* n ≤ 100
* -1000 ≤ xi, yi ≤ 1000
* 0 <ri ≤ 100

Input

Multiple datasets are given as input. Each dataset is given in the following format:

n (number of crop circles: integer)
x1 y1 r1 (center coordinates and radius of the first circle: real numbers separated by blanks)
x2 y2 r2 (center coordinates and radius of the second circle: real numbers separated by blanks)
..
..
xn yn rn (center coordinates and radius of the nth circle: real numbers separated by blanks)


When n is 0, it indicates the end of input.

Output

Output the string length on one line for each dataset. The output may contain an error of 0.000001 or less.

Example

Input

4
6 4 2
4 7 3
7 10 3
13 6 1
4
7 4 2
4 7 3
7 10 3
11 6 3
0


Output

39.664699289572
45.627024663706

Solution (in CPP):

Solution:
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;

using ld = double;
using Point =  complex<ld>;
const ld eps = 1e-9;
const ld pi = acos(-1.0);

bool eq(ld a, ld b) {
	return (abs(a - b) < eps);
}

bool cmp(Point x,Point y){
	if(eq(x.real(),y.real()))return x.imag()<y.imag();
	return x.real()<y.real();
}

bool eqq(Point x,Point y){
	return eq(x.real(),y.real())&&eq(x.imag(),y.imag());
}
//内積
ld dot(Point a, Point b) {
	return real(conj(a)*b);
}
//外積
ld cross(Point a, Point b) {
	return imag(conj(a)*b);
}


//線分
//直線にするなら十分通い２点を端点とすればよい
class Line {
public:
	Point a, b;
};
//円
class Circle {
public:
	Point p;
	ld r;
};
//3点の位置関係
int ccw(Point a, Point b, Point c) {
	b -= a; c -= a;
	if (cross(b, c) > eps)return 1;//a,b,cが反時計回り
	if (cross(b, c) < -eps)return -1;//a,b,cが時計回り
	if (dot(b, c) < 0)return 2;//c,a,bの順に一直線
	if (norm(b) < norm(c))return -2;//a,b,cの順に一直線
	return 0;//a,c,bの順に一直線
}
//2直線の交差判定
bool isis_ll(Line l, Line m) {
	return !eq(cross(l.b - l.a, m.b - m.a), 0);
}
//直線と線分の交差判定
bool isis_ls(Line l, Line s) {
	return (cross(l.b - l.a, s.a - l.a)*cross(l.b - l.a, s.b - l.a) < eps);
}
//点が直線上に存在するか
bool isis_lp(Line l, Point p) {
	return (abs(cross(l.b - p, l.a - p)) < eps);
}
//点が線分上に存在するか
bool isis_sp(Line s, Point p) {
	return (abs(s.a - p) + abs(s.b - p) - abs(s.b - s.a) < eps);
}
//線分と線分の交差判定
bool isis_ss(Line s, Line t) {
	if (isis_sp(s, t.a) || isis_sp(s, t.b) || isis_sp(t, s.a) || isis_sp(t, s.b))return true;
	return(cross(s.b - s.a, t.a - s.a)*cross(s.b - s.a, t.b - s.a) < -eps && cross(t.b - t.a, s.a - t.a)*cross(t.b - t.a, s.b - t.a) < -eps);
}
//点から直線への垂線の足
Point proj(Line l, Point p) {
	ld t = dot(p - l.a, l.a - l.b) / norm(l.a - l.b);
	return l.a + t * (l.a - l.b);
}


//直線と直線の交点
//平行な２直線に対しては使うな！！！！
Point is_ll(Line s, Line t) {
	Point sv = s.b - s.a; Point tv = t.b - t.a;
	return s.a + sv * cross(tv, t.a - s.a) / cross(tv, sv);
}
//直線と点の距離
ld dist_lp(Line l, Point p) {
	return abs(p - proj(l, p));
}
//直線と直線の距離
ld dist_ll(Line l, Line m) {
	return isis_ll(l, m) ? 0 : dist_lp(l, m.a);
}
//線分と直線の距離
ld dist_ls(Line l, Line s) {
	return isis_ls(l, s) ? 0 : min(dist_lp(l, s.a), dist_lp(l, s.b));
}
//線分と点の距離
ld dist_sp(Line s, Point p) {
	Point r = proj(s, p);
	return isis_sp(s, r) ? abs(p - r) : min(abs(p - s.a), abs(p - s.b));
}
//線分と線分の距離
ld dist_ss(Line s, Line t) {
	if (isis_ss(s, t))return 0;
	return min({ dist_sp(s,t.a),dist_sp(s,t.b),dist_sp(t,s.a),dist_sp(t,s.b) });
}

vector<Point> is_cc(Circle c1,Circle c2){
    if(c1.r<c2.r)swap(c1,c2);
    ld d = abs(c1.p-c2.p);
    if(c1.r+c2.r<=abs(c1.p-c2.p))return {};
    if(c1.r>=c2.r+abs(c1.p-c2.p))return {};
    Point vec = (c2.p-c1.p)/abs(c2.p-c1.p);
    ld theta = acos((c1.r*c1.r + d*d - c2.r*c2.r)/(2.0*c1.r*d));
    vector<Point> ret;
    ret.push_back(c1.p+c1.r*cos(theta)*vec+c1.r*sin(theta)*vec*Point(0,1));
    ret.push_back(c1.p+c1.r*cos(theta)*vec-c1.r*sin(theta)*vec*Point(0,1));
    return ret;
}


void solve(int n){
    Circle c[n];
    rep(i,n){
        ld x,y,r;
        cin>>x>>y>>r;
        c[i]={Point(x,y),r};
    }
    vector<ld> cross[n];
    rep(i,n)rep(j,i){
        auto res = is_cc(c[i],c[j]);
        for(auto p : res){
            cross[i].push_back(arg(p-c[i].p));
            cross[j].push_back(arg(p-c[j].p));
        }
    }
    ld ans = 0;
    rep(i,n){
        if(cross[i].empty()){
            bool ok=true;
            rep(k,n){
                    if(k==i)continue;
                    if(abs(c[i].p+c[i].r*Point(1,0)-c[k].p)<c[k].r)ok=false;
                }
            if(ok)ans += c[i].r * 2.0 * pi ;
        }

        else {
            sort(cross[i].begin(),cross[i].end());
            int m =cross[i].size();
            rep(j,m){
                ld t1=cross[i][j],t2 = cross[i][(j+1)%m];
                if(t1>t2)t2+=2.0*pi;
                Point mid = c[i].p + c[i].r * Point(cos((t1+t2)*.5),sin((t1+t2)*.5));
                bool ok=true;
                rep(k,n){
                    if(k==i)continue;
                    if(abs(mid-c[k].p)<=c[k].r)ok=false;
                }
                if(ok) ans += c[i].r*(t2-t1);
            }
        }
    }
    cout<<ans<<endl;
}

int main(){
   int n;
   cout<<fixed<<setprecision(10);
   while(cin>>n,n)solve(n);
    return 0;
}