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>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef double R;//double long double ??????????????? cmath?????¢??°??????????????????????????????????????????????????§????????????
typedef complex<R> Point;
typedef pair<Point , Point> Line;
typedef pair<Point ,R > Circle;
typedef vector<Point> Poly;

#define EPS (1e-10)//??????
#define EQ(a,b) (abs((a)-(b)) < EPS)//??????????????°???????????????
#define EQV(a,b) ( EQ((a).real(), (b).real()) && EQ((a).imag(), (b).imag()) )//????????????????????????????????????

#define pb push_back
int dy[]={0, 0, 1, -1, 0};
int dx[]={1, -1, 0, 0, 0};
 
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)

int n;
Circle c[200];

R dot(Point a,Point b){
    return (a.real() * b.real() + a.imag() * b.imag());
}
R cross(Point a,Point b){
    return (a.real() * b.imag() - a.imag() * b.real());
}
bool is_intersection_two_line_segment(Line a,Line b){
    Point a1 = a.first,a2 = a.second,b1 = b.first,b2 = b.second;
    return ( cross(a2 - a1,b1 - a1) * cross(a2 - a1,b2 - a1) < EPS) && (cross(b2 - b1,a1 - b1) * cross(b2 - b1, a2 - b1) < EPS);
}

Line intersection_of_two_circles(Circle c1,Circle c2){//????????????????????????Line?????\????????????(r1 + r2 > sqrt((x1 - x2) ^ 2 + (y1 - y2) ^ 2))???????????????????????????
    R a =  abs(c2.first - c1.first);
    R b = c1.second;
    R c = c2.second;

    R rc = (a  * a + b * b - c * c) / (2.0 * a);
    R rs = sqrt(b * b - rc * rc);//C++ ??§???????????????????????????????????§???????????????sqrt ????????? float ?????????????????? long double ????????????????????????????????????????????¨?????§???????????? C ????????°????????§??????sqrt ????????? double ??????????????£??????????????????
    Point diff = (c2.first - c1.first) / a;

    Line p ;
    p.first = c1.first + diff * rc + diff * Point(0,1) * rs;
    p.second = c1.first + diff * rc + diff * Point(0,-1) * rs;

    return p;
}

bool input(){
    scanf("%d",&n);

    if(n == 0)
        return false;
    REP(i,n){
        double re,im;
        cin >> re >> im  >> c[i].second;
        c[i].first = Point(re,im);
    }
    return true;
}

double G[1000][1000];
Point ver[200][2];

bool judgecross(Point a,Point b,int a1,int b1){
    Line c(a,b);
     FOR(i,a1+1,b1){
         Line d(ver[i][0],ver[i][1]);
         if(!is_intersection_two_line_segment(c,d))
             return false;
    }
    return true;
}


void solve(){
    REP(i,1000)
        REP(j,1000)
            G[i][j] = 1000000000.0;
    REP(i,1000)
        G[i][i] = 0.0;

    REP(i,n-1){
        Line inter = intersection_of_two_circles(c[i],c[i+1]);
        ver[i][0] = inter.first;
        ver[i][1] = inter.second;
    }
    //REP(i,n-1)
        //printf("%lf %lf    %lf %lf\n",ver[i][0].real(),ver[i][0].imag(),ver[i][1].real(),ver[i][1].imag());
    REP(i,(n - 1) * 2){
        FOR(j,i + 1,(n - 1) * 2){
            int i1 = i / 2;int i2 = i % 2;
            int j1 = j / 2;int j2 = j % 2;
            if(!judgecross(ver[i1][i2],ver[j1][j2],i1,j1))
                continue;
            G[i][j] = G[j][i] = abs(ver[i1][i2] - ver[j1][j2]);
        }
    }
    Point start(c[0].first.real(),c[0].first.imag());
    REP(i,(n - 1) * 2){
        int i1 = i / 2;int i2 = i % 2;
        if(!judgecross(start,ver[i1][i2],-1,i1))
            continue;
        G[i][n * 2 - 2]  = G[n * 2 - 2][i] = abs(start - ver[i1][i2]);
    }
    Point end(c[n-1].first.real(),c[n-1].first.imag());
    REP(i,(n - 1) * 2){
        int i1 = i / 2;int i2 = i % 2;
        if(!judgecross(ver[i1][i2],end,i1,n-1))
            continue;
        G[i][n * 2 - 1] = G[n * 2 - 1][i] = abs(end - ver[i1][i2]);
    }
    if(judgecross(start,end,-1,n-1))
        G[n * 2 - 1][n * 2 - 2] = G[n * 2 - 2][n * 2 - 1] = abs(start - end);
    REP(k,n*2){
        REP(i,n*2){
            REP(j,n*2){
                G[i][j] = min(G[i][j],G[i][k] + G[k][j]);
            }
        }
    }
    printf("%lf\n",G[n*2-2][n*2-1]);
}

int main(){
    while(input()){
        solve();
    }
    return 0;
}