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*/
#include<iostream>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<utility>
#include<vector>
#include<cmath>
#include<cstdio>
#include<complex>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define rp(a) while(a--)
#define pb push_back
#define mp make_pair
#define it ::iterator
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
const double PI=acos(-1);
const double EPS=1e-10;
const double inf=1e8;
using namespace std;
#define shosu(x) fixed<<setprecision(x)
typedef complex<double> P;
typedef vector<P> G;
typedef vector<int> vi;
typedef vector<vi> vvi;

struct L : public vector<P> {
  L(const P &a, const P &b) {
    push_back(a); push_back(b);
  }
};
struct C{
	P c;double r;
	C(const P &c,double r):c(c),r(r){}
};
#define curr(P, i) P[i]
#define next(P, i) P[(i+1)%P.size()]
#define diff(P, i) (next(P, i) - curr(P, i))
namespace std {
	bool operator < (const P& a, const P& b) {
		return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);
		//return imag(a) != imag(b) ? imag(a) < imag(b) : real(a) < real(b);  
	}
	bool operator == (const P& a, const P& b) {
		return a.real()==b.real()&&a.imag()==b.imag();
	}
}
P pin(){
	double x,y;
	char d;
	cin>>x>>y;
	P p(x,y);
	return p;
}
void PIN(P* a,int n){
	rep(i,n)a[i]=pin();
}
double dot(P a,P b){
	return real(conj(a)*b);
}
double cross(P a,P b){
	return imag(conj(a)*b);
}
int ccw(P a, P b, P c) {
  b -= a; c -= a;
  if (cross(b, c) > 0)   return +1;       // counter clockwise
  if (cross(b, c) < 0)   return -1;       // clockwise
  if (dot(b, c) < 0)     return +2;       // c--a--b on line
  if (norm(b) < norm(c)) return -2;       // a--b--c on line
  return 0;
}
P projection(L a,P p){
	double t=dot(p-a[0],a[0]-a[1])/norm(a[0]-a[1]);
	return a[0]+t*(a[0]-a[1]);
}
P reflection(L a,P p){
	return p+2.0*(projection(a,p)-p);
}
bool intersectLL(const L &l, const L &m) {
  return abs(cross(l[1]-l[0], m[1]-m[0])) > EPS || // non-parallel
         abs(cross(l[1]-l[0], m[0]-l[0])) < EPS;   // same line
}
bool intersectLS(const L &l, const L &s) {
  return cross(l[1]-l[0], s[0]-l[0])*       // s[0] is left of l
         cross(l[1]-l[0], s[1]-l[0]) < EPS; // s[1] is right of l
}
bool intersectLP(const L &l, const P &p) {
  return abs(cross(l[1]-p, l[0]-p)) < EPS;
}
bool intersectSS(const L &s, const L &t) {
  return ccw(s[0],s[1],t[0])*ccw(s[0],s[1],t[1]) <= 0 &&
         ccw(t[0],t[1],s[0])*ccw(t[0],t[1],s[1]) <= 0;
}
bool intersectSP(const L &s, const P &p) {
  return abs(s[0]-p)+abs(s[1]-p)-abs(s[1]-s[0]) < EPS; // triangle inequality
}
double distanceLP(const L &l, const P &p) {
	return abs(p - projection(l, p));
}
double distanceLL(const L &l, const L &m) {
  return intersectLL(l, m) ? 0 : distanceLP(l, m[0]);
}
double distanceLS(const L &l, const L &s) {
  if (intersectLS(l, s)) return 0;
  return min(distanceLP(l, s[0]), distanceLP(l, s[1]));
}
double distanceSP(const L &s, const P &p) {
  const P r = projection(s, p);
  if (intersectSP(s, r)) return abs(r - p);
  return min(abs(s[0] - p), abs(s[1] - p));
}
double distanceSS(const L &s, const L &t) {
  if (intersectSS(s, t)) return 0;
  return min(min(distanceSP(s, t[0]), distanceSP(s, t[1])),
             min(distanceSP(t, s[0]), distanceSP(t, s[1])));
}
/*bool intersectCS(C c,const L &l){
    return (distanceLP(l,c.c) < c.r+EPS &&
            (c.r < abs(c.c-l[0]) + EPS || c.r < abs(c.c-l[1]) + EPS));
}*/
int intersectCS(C c,L &l){
	if(norm(projection(l,c.c)-c.c)-c.r*c.r>EPS)return 0;
	const double d1=abs(c.c-l[0]),d2=abs(c.c-l[1]);
	if(d1<c.r+EPS&&d2<c.r+EPS)return 0;
	if(d1<c.r-EPS&&d2>c.r+EPS||d1>c.r+EPS&&d2<c.r-EPS)return 1;
	const P h=projection(l,c.c);
	if(dot(l[0]-h,l[1]-h)<0)return 2;
	return 0;
}
P crosspointSS(L a,L b){
	double t1=abs(cross(a[1]-a[0],b[0]-a[0]));
	double t2=abs(cross(a[1]-a[0],b[1]-a[0]));
	return b[0]+(b[1]-b[0])*t1/(t1+t2);
}
L crosspointCL(C c,L l){
	P pr=projection(l,c.c);
	P e=(l[1]-l[0])/abs(l[1]-l[0]);
	double t=sqrt(c.r*c.r-norm(pr-c.c));
	P a=pr+t*e;
	P b=pr-t*e;
	if(b<a)swap(a,b);
	return L(a,b);
}
L crosspointCS(C c,L l){
	if(intersectCS(c,l)==2)return crosspointCL(c,l);
	L ret=crosspointCL(c,l);
	if(dot(l[0]-ret[0],l[1]-ret[0])<0)ret[1]=ret[0];
	else ret[0]=ret[1];
	return ret;
}
L crosspointCC(C a,C b){
	P tmp=b.c-a.c;
	double d=abs(tmp);
	double q=acos((a.r*a.r+d*d-b.r*b.r)/(2*a.r*d));
	double t=arg(tmp);//atan(tmp.imag()/tmp.real());
	P p1=a.c+polar(a.r,t+q);
	P p2=a.c+polar(a.r,t-q);
	if(p2<p1)swap(p1,p2);
	return L(p1,p2);
}
P crosspointLL(const L &l, const L &m) {
  double A = cross(l[1] - l[0], m[1] - m[0]);
  double B = cross(l[1] - l[0], l[1] - m[0]);
  if (abs(A) < EPS && abs(B) < EPS) return m[0]; // same line
  return m[0] + B / A * (m[1] - m[0]);
}
double area(const G &g){
    double S =0;
    for(int i =0;i <g.size();i++){
        S +=(cross(g[i],g[(i+1)%g.size()]));
    }
    return abs(S/2.0);
}
bool isconvex(const G &g){
	int n=g.size();
	rep(i,n)if(ccw(g[(i+n-1)%n],g[i%n],g[(i+1)%n])==-1)return false;
	return true;
}
int inconvex(const G& g, const P& p) {
	bool in = false;
	int n=g.size();
	rep(i,n){
		P a=g[i%n]-p;
		P b=g[(i+1)%n]-p;
		if(imag(a)>imag(b))swap(a, b);
		if(imag(a)<=0&&0<imag(b))if(cross(a,b)<0)in=!in;
		if(cross(a,b)==0&&dot(a,b)<=0)return 1;//ON
	}
	return in?2:0;//IN : OUT;
}
G convex_hull(G &ps) {
    int n=ps.size(),k=0;
	sort(ps.begin(), ps.end());
	G ch(2*n);
	for(int i=0;i<n;ch[k++]=ps[i++])//lower-hull
		while(k>=2&&ccw(ch[k-2],ch[k-1],ps[i])==-1)--k;//<=0  ->  ==-1
	for(int i=n-2,t=k+1;i>=0;ch[k++]=ps[i--])//upper-hull
		while(k>=t&&ccw(ch[k-2],ch[k-1],ps[i])==-1)--k;//
	ch.resize(k-1);
	return ch;
}
double convex_diameter(const G &pt) {
  const int n = pt.size();
  int is = 0, js = 0;
  for (int i = 1; i < n; ++i) {
    if (imag(pt[i]) > imag(pt[is])) is = i;
    if (imag(pt[i]) < imag(pt[js])) js = i;
  }
  double maxd = norm(pt[is]-pt[js]);

  int i, maxi, j, maxj;
  i = maxi = is;
  j = maxj = js;
  do {
    if (cross(diff(pt,i), diff(pt,j)) >= 0) j = (j+1) % n;
    else i = (i+1) % n;
    if (norm(pt[i]-pt[j]) > maxd) {
      maxd = norm(pt[i]-pt[j]);
      maxi = i; maxj = j;
    }
  } while (i != is || j != js);
  return sqrt(maxd); /* farthest pair is (maxi, maxj). */
}//convex_diameter(g)
G convex_cut(const G& g, const L& l) {
  G Q;
  for (int i = 0; i < g.size(); ++i) {
    P a= curr(g, i), b= next(g, i);
    if (ccw(l[0], l[1], a) != -1) Q.push_back(a);
    if (ccw(l[0], l[1], a)*ccw(l[0], l[1], b) < 0)
      Q.push_back(crosspointLL(L(a,b), l));
  }
  return Q;
}
P turn(P p,double t){
	return p*exp(P(.0,t*PI/180.0));
}
P turn2(P p,double t){
	return p*exp(P(.0,t));
}
vector<L> tangentCC(C a,C b){
	if(a.r<b.r)swap(a,b);
	double d=abs(a.c-b.c);
	vector<L>l;
	if(d<EPS)return l;
	if(a.r+b.r<d-EPS){//hanareteiru
		double t=acos((a.r+b.r)/d);
		t=t*180/PI;
		l.pb(L(a.c+turn(a.r/d*(b.c-a.c),t),b.c+turn(b.r/d*(a.c-b.c),t)));
		l.pb(L(a.c+turn(a.r/d*(b.c-a.c),-t),b.c+turn(b.r/d*(a.c-b.c),-t)));
	}else if(a.r+b.r<d+EPS){//kuttuiteiru soto
		P p=a.c+a.r/d*(b.c-a.c);
		l.pb(L(p,p+turn(b.c-a.c,90)));
	}
	if(abs(a.r-b.r)<d-EPS){//majiwatteiru
		double t1=acos((a.r-b.r)/d);
		t1=t1*180/PI;
		double t2=180-t1;
		l.pb(L(a.c+turn(a.r/d*(b.c-a.c),t1),b.c+turn(b.r/d*(a.c-b.c),-t2)));
		l.pb(L(a.c+turn(a.r/d*(b.c-a.c),-t1),b.c+turn(b.r/d*(a.c-b.c),t2)));
	}else if(abs(a.r-b.r)<d+EPS){//kuttuiteiru uti
		P p=a.c+a.r/d*(b.c-a.c);
		l.pb(L(p,p+turn(b.c-a.c,90)));
	}
	return l;
}
void printL(const L &out){
	printf("%.9f %.9f %.9f %.9f\n",out[0].real(),out[0].imag(),out[1].real(),out[1].imag());
}
C CIN(){
	P p=pin();
	double r;
	cin>>r;
	return C(p,r);
}
bool para(L a,L b){
	return (abs(cross(a[1]-a[0],b[1]-b[0]))<EPS);
}
int main(){
	int n;
	while(cin>>n,n){
		vector<C>c;
		rep(i,n){
			double a,b,d;
			cin>>a>>b>>d;
			c.pb(C(P(a,b),d));
		}
		vector<C>tmp;
		rep(i,n){
			bool h=true;
			rep(j,n)if(i!=j&&abs(c[i].c-c[j].c)<c[j].r-c[i].r)
				h=false;
			if(h)tmp.pb(c[i]);
		}
		c=tmp;
		n=c.size();
		double out=0;
		rep(i,n){
			G g;
			rep(j,n)if(i!=j&&abs(c[i].c-c[j].c)<c[i].r+c[j].r){
				L l=crosspointCC(c[i],c[j]);
				g.pb(l[0]);
				if(l[0]!=l[1])g.pb(l[1]);
			}
			if(g.size()==0){
				out+=2*PI*c[i].r;
				continue;
			}
			rep(j,g.size()-1)rep(k,g.size()-j-1)if(arg(g[k]-c[i].c)>arg(g[k+1]-c[i].c)){
				swap(g[k+1],g[k]);
			}
			rep(j,g.size()){
				double t=arg(g[(j+1)%g.size()]-c[i].c)+arg(g[j]-c[i].c);
				P p=c[i].c+turn2(P(c[i].r,0),t/2+(j==g.size()-1?PI:0));
				bool h=true;
				rep(k,n)if(i!=k&&abs(c[k].c-p)<c[k].r){
					h=false;
					break;
				}
				t=abs(arg(g[(j+1)%g.size()]-c[i].c)-arg(g[j]-c[i].c));
				if(j==g.size()-1)t=2*PI-t;
				if(h)out+=2*c[i].r*t/2;
//				cout<<i<<" "<<j<<" "<<h<<endl;
			}
		}
		printf("%.9f\n",out);
	}
}