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
#include <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <sstream>
#include <map>
#include <set>
#include <cmath>
#include <queue>

#define REP(i,k,n) for(int i=k;i<n;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1<<30
#define pb push_back
#define mp make_pair
#define EPS 1e-8
#define equals(a,b) fabs((a) - (b)) < EPS

using namespace std;
typedef long long ll;
typedef pair<int,int> P;

struct Point {
	double x, y;

	Point(double x=0, double y=0) : x(x), y(y) {}

	Point operator+(const Point &o) const { return Point(x+o.x, y+o.y); }

	Point operator-(const Point &o) const { return Point(x-o.x, y-o.y); }

	Point operator*(const double m) const { return Point(x*m, y*m); }

	Point operator/(const double d) const { return Point(x/d, y/d); }

	bool operator<(const Point &o) const { return x != o.x ? x < o.x : y < o.y; }

	bool operator==(const Point &o) const { return fabs(x-o.x) < EPS && fabs(y-o.y) < EPS; }

	double cross(const Point &o) const { return x * o.y - y * o.x; }

	double dot(const Point &o) const { return x * o.x + y * o.y; }

	double atan() const { return atan2(y, x); }

	double norm() const { return sqrt(dot(*this)); }

	double distance(const Point &o) const { return (o - (*this)).norm(); }

	double area(const Point &a,const Point &b) {
		Point p = a - (*this), p2 = b - (*this); 
		return p.cross(p2);
	}

	double area_abs(const Point &a,const Point &b) const {
		Point p = a - (*this), p2 = b - (*this);
		return fabs(p.cross(p2)) / 2.0;
	}	

	//線分abが自身に含まれているのかどうか判断する
	int between(const Point &a,const Point &b) {
		if(area(a,b) != 0) return 0;

		if(a.x != b.x)  return ((a.x <= x) && (x <= b.x)) || ((a.x >= x) && (x >= b.x));
		else return ((a.y <= y) && (y <= b.y)) || ((a.y >= y) && (y >= b.y));
	}      

	double distance_seg(const Point& a,const Point& b) {
		if((b-a).dot(*this-a) < EPS) {
			return (*this-a).norm();
		}
		if((a-b).dot(*this-b) < EPS) {
			return (*this-b).norm();
		}
		return abs((b-a).cross(*this-a)) / (b-a).norm();
	}

	bool hitPolygon(const Point& a,const Point& b,const Point& c) {
		double t = (b-a).cross(*this-b);
		double t2 = (c-b).cross(*this-c);
		double t3 = (a-c).cross(*this-a);	

		if((t > 0 && t2 > 0 && t3 > 0) || ( t < 0 && t2 < 0 && t3 < 0)) {
			return true;
		}

		return false;
	}
};

istream &operator >> (istream &is, Point &p) {
	return is >> p.x >> p.y;
}

ostream &operator << (ostream &out, const Point &p) {
	cout << "(" << p.x << "," << p.y << ")";
	return out;
}

struct Seg {
	Point a,b;

	Seg (Point a, Point b) : a(a),b(b) {}

	bool isOrthogonal(Seg &s) { return equals((b - a).dot(s.b - s.a),0.0); }

	bool isParallel(Seg &s) { return equals((b-a).cross(s.b - s.a),0.0); }

	bool isIntersect(Seg &s) {
		if(s.a.between(a,b) || s.b.between(a,b) || a.between(s.a,s.b) || b.between(s.a,s.b)) {
			return true;
		}
		return ((a-b).cross(s.a-a) * (a-b).cross(s.b-a) < EPS) && ((s.b-s.a).cross(a-s.a)*(s.b-s.a).cross(b-s.a) < EPS);
	}

	bool distance(Seg &s) {
		if((*this).isIntersect(s)) return 0.0;

		return min(min(a.distance_seg(s.a,s.b),b.distance_seg(s.a,s.b)),min(s.a.distance_seg(a,b),s.b.distance_seg(a,b)));
	}

	Point getCrossPoint(Seg &s) {
		Point p = s.b - s.a;
		double d = abs(p.cross(a-s.a));
		double d2 = abs(p.cross(b-s.a));

		double t = d / (d+d2);
		return a + (b-a)*t;
	}

	Point project(Point &p) {
		Point base = b - a;
		double t = base.dot(p-a) / base.dot(base);
		return a + base * t;
	}

	Point reflect(Point &p) {
		return p + (project(p) - p) * 2.0;
	}
};

struct Circle {
	Point p;
	double r;

	Circle() : p(Point(0,0)), r(0) {}

	Circle(Point o, double r) : p(o), r(r) {}

	Circle(double x,double y, double r) : p(Point(x,y)), r(r) {}

	bool isCircleIn(const Point& o) {
		Point res = o-p;
		return res.dot(res) < r*r + EPS;
	}

	// 1:外で接する，0:交差なし，-1:内で接する，2:交差，-2:内包
	int isIntersect(const Circle& c) {
		double d = (c.p - p).dot(c.p - p);
		double len = (c.r + r) * (c.r + r);

		if(equals(d,len)) return 1;
		if(d > len) return 0;

		double R = fabs(c.r - r) * fabs(c.r - r);
		if(equals(d,R)) return -1;
		if(d > R) return 2;
		return -2;
	}

	vector<Point> getCrossPoint(const Circle& c) {
		vector<Point> ret;
		int ch = isIntersect(c);

		if(ch == 0 || ch == -2) return ret;

		Point base = c.p - p;
		double len = base.dot(base);
		double t = (r*r - c.r*c.r + len) / (2.0 * len);

		if(ch == 2) {
			Point n(-base.y,base.x);
			n = n / (n.norm());
			double h = sqrt(r * r - t*t*len);

			ret.push_back(p + (base*t) + (n*h));
			ret.push_back(p + (base*t) - (n*h));
		} else {
			ret.push_back(p + (base*t));
		}

		return ret;
	}
};

struct edge {
	int from,to;
	double cost;

	edge(int t,double c) : to(t),cost(c) {}
	edge(int f,int t,double c) : from(f),to(t),cost(c) {}

	bool operator<(const edge &e) const {
		return cost < e.cost;
	}
};

vector<edge> G[100 * 100 + 5];
double d[100 * 100 + 5];

void dijkstra(int s,int n) {
	priority_queue<P,vector<P>,greater<P> > que;
	fill(d,d+n,INF);

	d[s] = 0;
	que.push(P(0,s));

	while(que.size()) {
		P p = que.top();
		que.pop();

		int v = p.second;
		if(d[v] < p.first) continue;

		rep(i,G[v].size()) {
			edge e = G[v][i];
			if(d[e.to] > d[v] + e.cost) {
				d[e.to] = d[v] + e.cost;
				que.push(P(d[e.to],e.to));
			}
		}
	}
}

int main() {
	int n;

	while(cin >> n && n) {
		rep(i, 100 * 100 + 5) {
			G[i].clear();
		}

		vector<Circle> v(n);
		rep(i, n) {
			cin >> v[i].p >> v[i].r;
		}

		vector<Seg> seg;
		rep(i, n - 1) {
			vector<Point> ret = v[i].getCrossPoint(v[i+1]);
			seg.push_back(Seg(ret[0], ret[1]));
		}

		rep(i, n-1) {
			Seg s1(v[0].p, seg[i].a);
			int id = (i+1) * 2;
			bool flag = true;

			rep(j, i) {
				if(s1.isIntersect(seg[j])) continue;
				flag = false;
			}

			if(flag) {
				double dist = v[0].p.distance(s1.b);
				G[0].push_back(edge(id-1, dist));
				G[id-1].push_back(edge(0, dist));
			}

			Seg s2(v[0].p, seg[i].b);
			flag = true;
			rep(j, i) {
				if(s2.isIntersect(seg[j])) continue;
				flag = false;
			}
			
			if(flag) {
				double dist = v[0].p.distance(s2.b);
				G[0].push_back(edge(id, dist));
				G[id].push_back(edge(0, dist));
			}
		}

		rep(i, n-1) {
			bool flag = true;
			int id = (i+1) * 2;

			REP(j, i+1, n-1) {
				Seg sj1(seg[i].a, seg[j].a);
				int nid = (j+1) * 2;
				flag = true;
				REP(k, i+1, j) {
					if(sj1.isIntersect(seg[k])) continue;
					flag = false;
				}
		
				if(flag) {
					double dist = seg[i].a.distance(seg[j].a);
					G[id-1].push_back(edge(nid-1, dist));
					G[nid-1].push_back(edge(id-1, dist));
				}
		
				Seg sj2(seg[i].a, seg[j].b);
				flag = true;
				REP(k, i+1, j) {
					if(sj2.isIntersect(seg[k])) continue;
					flag = false;
				}
		
				if(flag) {
					double dist = seg[i].a.distance(seg[j].b);
					G[id-1].push_back(edge(nid, dist));
					G[nid].push_back(edge(id-1, dist));
				}
		
				Seg sj3(seg[i].b, seg[j].a);
				flag = true;
				REP(k, i+1, j) {
					if(sj3.isIntersect(seg[k])) continue;
					flag = false;
				}
		
				if(flag) {
					double dist = seg[i].b.distance(seg[j].a);
					G[id].push_back(edge(nid-1, dist));
					G[nid-1].push_back(edge(id, dist));
				}
		
				Seg sj4(seg[i].b, seg[j].b);
				flag = true;
				REP(k, i+1, j) {
					if(sj4.isIntersect(seg[k])) continue;
					flag = false;
				}
		
				if(flag) {
					double dist = seg[i].b.distance(seg[j].b);
					G[id].push_back(edge(nid, dist));
					G[nid].push_back(edge(id, dist));
				}
			}
		
		}

		rep(i, n-1) {
			Seg s3(seg[i].a, v[n-1].p);
			bool flag = true;
			int id = (i+1) * 2;

			REP(j, i+1, n-1) {
				if(s3.isIntersect(seg[j])) continue;
				flag = false;
			}

			if(flag) {
				double dist = seg[i].a.distance(v[n-1].p);
				G[n*2-1].push_back(edge(id-1, dist));
				G[id-1].push_back(edge(n*2-1, dist));
			}

			Seg s4(seg[i].b, v[n-1].p);
			flag = true;
			REP(j, i+1, n-1) {
				if(s4.isIntersect(seg[j])) continue;
				flag = false;
			}

			if(flag) {
				double dist = seg[i].b.distance(v[n-1].p);
				G[n*2-1].push_back(edge(id, dist));
				G[id].push_back(edge(n*2-1, dist));
			}
		}

		Seg s(v[0].p, v[n-1].p);
		bool flag = true;
		rep(i, n-1) {
			if(s.isIntersect(seg[i])) continue;
			flag = false;
		}

		if(flag) {
			double dist = v[0].p.distance(v[n-1].p);
			G[0].push_back(edge(n*2-1, dist));
			G[n*2-1].push_back(edge(0, dist));
		}

		dijkstra(0, n*2);

		cout << fixed;
		cout.precision(10);

		cout << d[n*2-1] << endl;
	}

	
	return 0;
}