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 <cstdio>
#include <string>
#include <vector>
#include <cmath>
#include <cstring>
#include <string>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <algorithm>
#include <stack>
#include <sstream>
  
using namespace std;
  
#define EPS 1e-9
 
struct point{
    double x, y;
    point(){x = 0;y = 0;}
    point(double a, double b) : x(a), y(b) {}
};
  
double dis(point p1,point p2){
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
 
int ccw(point p1, point p2, point p3) {
    double ret = (p2.x - p1.x)*(p3.y - p1.y) - (p2.y - p1.y)*(p3.x - p1.x);
    if(abs(ret)<EPS) return 0;
    return ret > 0 ? 1 : -1;
}
 
struct Line {
	point p1,p2;
	double A,B,C;
    double getDis() {
        return dis(p1,p2);
    }
    Line() {}
    Line(double xx1, double yy1, double xx2, double yy2) {
		p1 = point(xx1,yy1), p2 = point(xx2,yy2);
		A = p1.y-p2.y, B = p2.x-p1.x, C = -(A*p1.x+B*p1.y);
	}
	Line(point t1, point t2) {
		p1 = t1, p2 = t2;
		A = p1.y-p2.y, B = p2.x-p1.x, C = -(A*p1.x+B*p1.y);
	}
};
  
bool is_in_line_seg(double x, double y, Line LL) {
    if(!( min(LL.p1.x,LL.p2.x)-EPS <= x && x <= max(LL.p1.x,LL.p2.x)+EPS 
        && min(LL.p1.y,LL.p2.y)-EPS <= y && y <= max(LL.p1.y,LL.p2.y)+EPS))
        return false;
    return abs( LL.A*x + LL.B*y + LL.C ) < EPS;
}

bool intersection_line_line(Line L1, Line L2, point &p) {
	double det = L1.A*L2.B - L2.A*L1.B;
	if(abs(det) < EPS) return 0;
	p.x = -(L2.B*L1.C - L1.B*L2.C) / det;
	p.y = -(L1.A*L2.C - L2.A*L1.C) / det;
	return 1;
}

bool is_cross_line_seg(Line L1, Line L2) {
	point p;
	if(!intersection_line_line(L1,L2,p)) return 0;
	return is_in_line_seg(p.x,p.y,L1) && is_in_line_seg(p.x,p.y,L2);
}
  
point intersection(point &u1,point &u2,point &v1,point &v2){
    point ret=u1;
    double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))
    /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
    ret.x+=(u2.x-u1.x)*t;
    ret.y+=(u2.y-u1.y)*t;
    return ret;
}
  
void intersection_line_circle(point c, double r,point &l1,point &l2,point& p1,point&p2){
    point p=c;
    double t;
    p.x+=l1.y-l2.y;
    p.y+=l2.x-l1.x;
    p=intersection(p,c,l1,l2);
    t=sqrt(r*r-dis(p,c)*dis(p,c))/dis(l1,l2);
    p1.x=p.x+(l2.x-l1.x)*t;
    p1.y=p.y+(l2.y-l1.y)*t;
    p2.x=p.x-(l2.x-l1.x)*t;
    p2.y=p.y-(l2.y-l1.y)*t;
}
  
void intersection_circle_circle(point c1, double r1,point c2,double r2,point& p1, point& p2){
    point u,v;
    double t;
    t=(1+(r1*r1-r2*r2)/dis(c1,c2)/dis(c1,c2))/2;
    u.x=c1.x+(c2.x-c1.x)*t;
    u.y=c1.y+(c2.y-c1.y)*t;
    v.x=u.x+c1.y-c2.y;
    v.y=u.y-c1.x+c2.x;
    intersection_line_circle(c1,r1,u,v,p1,p2);
}

point P[111],cP[111][2];
int X[111],Y[111],R[111],n;
double dp[111][2];
Line LL[111];
 
int main() {
#ifdef LOCAL
	freopen("input.txt","r",stdin);
#endif

    while(1) {
        for(int i=0;i<111;++i) dp[i][0] = 1e20, dp[i][1] = 1e20;
        scanf("%d",&n);
        if(n==0) break;
        for(int i=0;i<n;++i) scanf("%d%d%d",X+i,Y+i,R+i), P[i] = point(X[i],Y[i]);
        cP[0][0] = P[0], cP[0][1] = P[0];
        cP[n][0] = P[n-1], cP[n][1] = P[n-1];
        for(int i=0;i<n-1;++i) {
            intersection_circle_circle(P[i],R[i],P[i+1],R[i+1],cP[i+1][0],cP[i+1][1]);
            LL[i+1] = Line(cP[i+1][0],cP[i+1][1]);
        }
 
        dp[0][0] = dp[0][1] = 0;
        for(int i=1;i<=n;++i) for(int ai=0;ai<2;++ai) {
            for(int j=0;j<i;++j) for(int aj=0;aj<2;++aj) {
                bool can = 1;
                Line line = Line(cP[i][ai],cP[j][aj]);
                for(int k=j+1;k<i;++k) {
                    if(!is_cross_line_seg(line,LL[k])) {can = 0; break;}
                }
 
                if(can) dp[i][ai] = min(dp[i][ai], dp[j][aj] + line.getDis());
            }
        } printf("%.10lf\n",dp[n][0]);
    }
}