Problem: There is a chain consisting of multiple circles on a plane. The first (last) circle of the chain only intersects with the next (previous) circle, and each intermediate circle intersects only with the two neighboring circles.

Your task is to find the shortest path that satisfies the following conditions.

* The path connects the centers of the first circle and the last circle.
* The path is confined in the chain, that is, all the points on the path are located within or on at least one of the circles.

Figure E-1 shows an example of such a chain and the corresponding shortest path.

<image>

Figure E-1: An example chain and the corresponding shortest path


Input

The input consists of multiple datasets. Each dataset represents the shape of a chain in the following format.

> n
>  x1 y1 r1
>  x2 y2 r2
>  ...
>  xn yn rn
>

The first line of a dataset contains an integer n (3 ≤ n ≤ 100) representing the number of the circles. Each of the following n lines contains three integers separated by a single space. (xi, yi) and ri represent the center position and the radius of the i-th circle Ci. You can assume that 0 ≤ xi ≤ 1000, 0 ≤ yi ≤ 1000, and 1 ≤ ri ≤ 25.

You can assume that Ci and Ci+1 (1 ≤ i ≤ n−1) intersect at two separate points. When j ≥ i+2, Ci and Cj are apart and either of them does not contain the other. In addition, you can assume that any circle does not contain the center of any other circle.

The end of the input is indicated by a line containing a zero.

Figure E-1 corresponds to the first dataset of Sample Input below. Figure E-2 shows the shortest paths for the subsequent datasets of Sample Input.

<image>

Figure E-2: Example chains and the corresponding shortest paths


Output

For each dataset, output a single line containing the length of the shortest chain-confined path between the centers of the first circle and the last circle. The value should not have an error greater than 0.001. No extra characters should appear in the output.

Sample Input


10
802 0 10
814 0 4
820 1 4
826 1 4
832 3 5
838 5 5
845 7 3
849 10 3
853 14 4
857 18 3
3
0 0 5
8 0 5
8 8 5
3
0 0 5
7 3 6
16 0 5
9
0 3 5
8 0 8
19 2 8
23 14 6
23 21 6
23 28 6
19 40 8
8 42 8
0 39 5
11
0 0 5
8 0 5
18 8 10
8 16 5
0 16 5
0 24 5
3 32 5
10 32 5
17 28 8
27 25 3
30 18 5
0


Output for the Sample Input


58.953437
11.414214
16.0
61.874812
63.195179






Example

Input

10
802 0 10
814 0 4
820 1 4
826 1 4
832 3 5
838 5 5
845 7 3
849 10 3
853 14 4
857 18 3
3
0 0 5
8 0 5
8 8 5
3
0 0 5
7 3 6
16 0 5
9
0 3 5
8 0 8
19 2 8
23 14 6
23 21 6
23 28 6
19 40 8
8 42 8
0 39 5
11
0 0 5
8 0 5
18 8 10
8 16 5
0 16 5
0 24 5
3 32 5
10 32 5
17 28 8
27 25 3
30 18 5
0


Output

58.953437
11.414214
16.0
61.874812
63.195179

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include <map>
//#include <utility>
#include <set>
#include <iostream>
//#include <memory>
#include <string>
#include <vector>
#include <algorithm>
//#include <functional>
#include <sstream>
//#include <deque>
#include <complex>
#include <stack>
#include <queue>
#include <cstdio>
//#include <cctype>
#include <cstring>
//#include <ctime>
#include <iterator>
#include <bitset>
//#include <numeric>
//#include <list>
//#include <iomanip>

#if __cplusplus >= 201103L
#include <array>
#include <tuple>
#include <initializer_list>
#include <unordered_set>
#include <unordered_map>
#include <forward_list>

#define cauto const auto&
#else

#endif

using namespace std;


namespace{
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

typedef vector<int> vint;
typedef vector<vector<int> > vvint;
typedef vector<long long> vll, vLL;
typedef vector<vector<long long> > vvll, vvLL;

#define VV(T) vector<vector< T > >

template <class T>
void initvv(vector<vector<T> > &v, int a, int b, const T &t = T()){
    v.assign(a, vector<T>(b, t));
}

template <class F, class T>
void convert(const F &f, T &t){
    stringstream ss;
    ss << f;
    ss >> t;
}


#define reep(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n) reep((i),0,(n))
#define ALL(v) (v).begin(),(v).end()
#define PB push_back
#define fi first
#define se second
#define mkp make_pair
#define RALL(v) (v).rbegin(),(v).rend()


#define MOD 1000000007LL
// #define EPS 1e-8
// static const int INF=1<<24;
const double EPS = 1e-8;
const double INF = 1e12;
const double PI = asin(0.5)*6;

typedef complex<double> P;

#define X real()
#define Y imag()

#define Curr(P,i) P[(i)%P.size()]
#define Next(P,i) P[(i+1)%P.size()]
#define Prev(P,i) P[(i+P.size()-1)%P.size()]

namespace std{
  bool operator<(const P a,const P b){
    return a.X != b.X ? a.X < b.X : a.Y < b.Y;
  }
}
double cross(const P a,const P b){
  return (conj(a)*b).imag();
}
double dot(const P a,const P b){
  return (conj(a)*b).real();
}
// TODO make graph (20)
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
  if(norm(b)<norm(c)) return -2;// a--b--c
                      return  0;// a--c--b(or b==c)
}

struct L : public vector<P>{
  L(){}
  L(const P a,const P b){
    push_back(a),push_back(b);
  }
};
typedef L S;
typedef vector<P> G;

struct C{
  P p;double r;
  C(){}
  C(const P p,double r): p(p),r(r){}
};



//直線と点の関係
//点から線分に垂線を下ろした点
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]);
}
//線対称の点
P reflection(L a,P p){
  return p + 2.0 * (projection(a,p)-p);
}




//交差判定
//TODO CP(内包) CL CS (0)
bool isCrossLL(L a,L b){
  return 
    abs(cross(a[1]-a[0],b[1]-b[0])) > EPS
    || abs(cross(a[1]-a[0],b[0]-a[0])) < EPS ;
}
bool isCrossLS(L a,S b){
  return 
    cross(a[1]-a[0],b[0]-a[0]) 
    * cross(a[1]-a[0],b[1]-a[0]) < EPS;
}
bool isCrossLP(L l,P p){
  return abs(cross(l[1]-p,l[0]-p)) < EPS;
}
bool isCrossSS(S a,S b){
  return
    ccw(a[0],a[1],b[0])
    * ccw(a[0],a[1],b[1]) <= 0
    &&
    ccw(b[0],b[1],a[0])
    * ccw(b[0],b[1],a[1]) <= 0;
}
bool isCrossSP(S a,P p){
  return abs(a[0]-p)+abs(a[1]-p)-abs(a[0]-a[1]) < EPS;
}


// 距離 CP CL CS は(distXP(x,c.p)-c.r)
double distPP(P a,P b){
  return abs(a-b);
}
double distLP(L a,P p){
  return abs(p-projection(a,p));
}
double distLL(L a,L b){
  return isCrossLL(a,b) ? 0 : distLP(a,b[0]);
}
double distLS(L a,S b){
  return isCrossLS(a,b) ? 0 : min(distLP(a,b[0]),distLP(a,b[1]));
}
double distSP(S a,P p){
  const P r = projection(a,p);
  return isCrossSP(a,r) ? abs(p-r) : min(abs(a[0]-p),abs(a[1]-p));
}
double distSS(S a,S b){
  return isCrossSS(a,b)?0:
    min(
      min(distSP(a,b[0]),distSP(a,b[1])),
      min(distSP(b,a[0]),distSP(b,a[1]))
    );
}

//円の交差判定
bool isCrossCP(C a,P p){
  return abs(a.p-p)-a.r<=EPS;
}
bool isCrossCL(C a,L l){
  return distLP(l,a.p)-a.r<EPS;
}
double distSP_MAX(S a,P p){
  return max(abs(a[0]-p),abs(a[1]-p));
}
bool isCrossCS(C a,S s){
  return distSP(s,a.p)-a.r<-EPS&&distSP_MAX(s,a.p)-a.r>+EPS;
}
bool isCrossCC(C a,C b){//接してる時は交差
  return abs(a.p-b.p)-(a.r+b.r) <= EPS;
}


//交差点
//先に交差判定をすること
P crossP_LL(L a,L b){
  double A = cross(a[1]-a[0],b[1]-b[0]);
  double B = cross(a[1]-a[0],a[1]-b[0]);
  if(abs(A)<EPS && abs(B)<EPS)return b[0];
  if(abs(A)<EPS)assert(false);
  return b[0]+B/A*(b[1]-b[0]);
}
vector<P> crossP_CL(C c,L l){
  P tmp = projection(l,c.p);
  P e = (l[0]-l[1])/abs(l[0]-l[1]);
  double h = abs(c.p-tmp)*abs(c.p-tmp);
  double t = sqrt(c.r*c.r - h*h);
  if(t<EPS)return {tmp};
  return {tmp + e*t,tmp - e*t};
}
vector<P> crossP_CC(C a,C b){
  P A = conj(b.p-a.p);
  P B = (b.r*b.r - a.r*a.r - (b.p-a.p)*conj(b.p-a.p));
  P C = a.r*a.r*(b.p-a.p);
  P D = B*B-4.0*A*C;
  P z1 = (-B+sqrt(D))/(2.0*A)+a.p;
  P z2 = (-B-sqrt(D))/(2.0*A)+a.p;
  return {z1,z2};
}



//三点->円
P PPPtoC(P a,P b,P c){
  P x = 1.0/(conj(b-a));
  P y = 1.0/(conj(c-a));
  return (y-x)/( conj(x)*y - x*conj(y) ) + a;
}

//凸包
G convex_hull(G ps){
  int n = ps.size();
  int k = 0;
  G ch(2*n);
  for(int i = 0;i<n;(ch[k++]=ps[i++])){
    while(k>=2&&ccw(ch[k-2],ch[k-1],ps[i])<=0)--k;
  }
  for(int i = n-2,t=k+1 ; i>=0 ; ch[k++]=ps[i--]){
    while(k>=t && ccw(ch[k-2],ch[k-1],ps[i])<=0)--k;
  }
  ch.resize(k-1);
  return ch;
} 


//凸性判定
bool isConvex(G g){
  for(int i=0; i<g.size();i++){
    if(ccw(Prev(g,i),Curr(g,i),Next(g,i))>0)return false;
  }
}



//接線
//TODO check
vector<L> TLine_CP(C c,P p){
  P v = c.p - p;
  double t = asin(abs(c.r)/(abs(v)));
  P e = v/abs(v) * exp(P(.0,t));
  P n1 = sqrt(abs(v)*abs(v) - c.r*c.r)*e + p;
  P n2 = reflection(L(p,c.p),n1);
  return {L(p,n1),L(p,n2)};
}

// TLine CC
vector<L> TLine_CPr(C c,P p,double r){
  P v = c.p - p;
  double t = asin(abs(c.r)/(abs(v)));
  P e = v/abs(v) * exp(P(.0,t));
  P n1 = sqrt(abs(v)*abs(v) - c.r*c.r)*e + p;
  P e1 = (n1-c.p)/abs(n1-c.p) * r;
  P n2 = reflection(L(p,c.p),n1);
  P e2 = (n2-c.p)/abs(n2-c.p) * r;
  return {L(p+e1,n1+e1),L(p+e2,n2+e2)};
}
vector<L> TLine_CC(C a,C b){
  //接してる時がヤバイ
  vector<L> res;
  if(!isCrossCC(a,b)&&(abs(a.r)>EPS)&&(abs(b.r>EPS))){
    P tmp = (a.p-b.p)*(b.r)/(a.r+b.r) + b.p;
    auto t1 = TLine_CP(a,tmp);
    auto t2 = TLine_CP(b,tmp);
    res.push_back(L(t1[0][1],t2[0][1]));
    res.push_back(L(t1[1][1],t2[1][1]));
  }
  if(abs(a.r-b.r)<EPS){
    const auto r = a.r;
    P e = (a.p-b.p)/abs(a.p-b.p) * exp(P(.0,90.0/180.0*PI));
    res.push_back(L(a.p+(e*r),b.p+(e*r)));
    if(abs(r)>=EPS)res.push_back(L(a.p-(e*r),b.p-(e*r)));
  }else{
    if(a.r<b.r)swap(a,b);
    auto t3 = TLine_CPr(C(a.p,a.r-b.r),b.p,b.r);
    for(auto i:t3){
      res.push_back(i);
    }
  }
  return res;
}

//原点を起点とした回転
P rotate(P a,double r){
  return a*exp(P(0.0,r*PI/180.0));
}

double dp[110][2];

void mainmain(){
	int n;
	while(cin>>n,n){
		vector<C> cir(n);
		rep(i,n){
			double a,b;
			cin>>a>>b>>cir[i].r;
      cir[i].p=P(a,b);
      // cout<<cir[i].p.X<<" "<<cir[i].p.Y<<" "<<cir[i].r<<endl;
		}
    VV(P) po;
    po.PB(vector<P>(1,cir[0].p));
    rep(i,n-1){
      po.PB(crossP_CC(cir[i],cir[i+1]));
    }
    po.PB(vector<P>(1,cir[n-1].p));
    rep(i,110){
      rep(j,2){
        dp[i][j]=INF;
      }
    }
    // printf("%.6lf %.6lf\n",po[1][0].X,po[1][0].Y);
    // printf("%.6lf %.6lf\n",po[1][1].X,po[1][1].Y);
    dp[0][0]=dp[0][1]=0;
    reep(i,1,n){
      if(i==1){
        dp[1][0]=distPP(cir[0].p,po[1][0]);
        dp[1][1]=distPP(cir[0].p,po[1][1]);
      }
      // dp[i][0]=mkp(distPP(cir[0].p,po[0][0]),0);
      else{
        rep(j,2){
          for(int k=i-1;k>=0;k--){
            rep(g,po[k].size()){
              S t1=S(po[i][j],po[k][g]);
              bool f=true;
              reep(h,k+1,i){
                f&=isCrossSS(t1,S(po[h][0],po[h][1]));
              }
              if(f){
                if(dp[i][j]>distPP(po[i][j],po[k][g])+dp[k][g]){
                  dp[i][j]=distPP(po[i][j],po[k][g])+dp[k][g];
                }
              }
            }
          }
        }
      }
    }
    double ans=INF;
    rep(i,n){
      rep(j,po[i].size()){
        // cout<<i<<" "<<j<<" ";
        // printf("%.6lf\n",dp[i][j]);
        S t1=S(po[i][j],po[n][0]);
        bool f=true;
        reep(k,i+1,n){
          f&=isCrossSS(t1,S(po[k][0],po[k][1]));
        }
        if(f){
          ans=min(ans,dp[i][j]+distPP(po[i][j],po[n][0]));
        }
      }
    }
    printf("%.6lf\n",ans);
    // rep(i,n){
      // cout<<dp[i][0].se<<" "<<dp[i][1].se<<endl;
    // }
	}
}



}
int main() try{
    mainmain();
}
catch(...){
}