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 <cstdio>
#include <complex>
#include <cmath>
#include <algorithm>
#include <vector>
#define rep(i, n) for(i = 0; i < n; i++)
#define double long double
using namespace std;
typedef complex<double> Point;
typedef vector<Point> Points;

int n;
Point c[100]; double r[100];

double dot(Point a, Point b) {
	return a.real() * b.real() + a.imag() * b.imag();
}

double cross(Point a, Point b) {
	return a.real() * b.imag() - a.imag() * b.real();
}

//a->b->cの曲がる方向。反時計回り, 180°折り返しは正.
const double EPS = 1e-10;		//大体, 座標範囲^2 / 10^(-16)くらいにしておくと無難？
//それぞれの条件を厳しめにしておくと, 線分交差判定はちょっと緩めにYesを返すようになったりして吉.
int ccw(Point a, Point b, Point c) {
	b -= a;
	c -= a;
	
	if (cross(b, c) > EPS) return +1;
	if (cross(b, c) < -EPS) return -1;
	if (dot(b, c) < -EPS) return +2;		//c--b--a
	if (norm(b) < norm(c) - EPS) return -2;	//a--b--c
	return 0;
}

//線分の交差判定. ギリギリ重ならない場合も重なると判定しておく.
bool intersect_SS(Point a, Point b, Point c, Point d) {
	return ccw(a, b, c) * ccw(a, b, d) <= 0 && ccw(c, d, a) * ccw(c, d, b) <= 0;
}

//円と直線の交点. 相異なる2点が交点になると仮定.
vector<Point> kouten_CL(Point c, double r, Point s, Point e) {
	s -= c;
	e -= c;
	
	double dt = dot(s, e - s);
	double nm = norm(e - s);
	double D = sqrt(dt * dt - nm * (norm(s) - r * r));
	double alpha1 = (-dt - D) / nm;
	double alpha2 = (-dt + D) / nm;
	
	Point res1 = s + alpha1 * (e - s);
	Point res2 = s + alpha2 * (e - s);
	vector<Point> ret;
	ret.push_back(res1 + c);
	ret.push_back(res2 + c);
	return ret;
}

//円と円の交点. 相異なる2点が交点になると仮定.
vector<Point> kouten_CC(Point c1, double r1, Point c2, double r2) {
	Point r = c2 - c1;
	double theta = acos((norm(r) + r1 * r1 - r2 * r2) / (2 * r1 * abs(r)));
	Point R = (r1 / abs(r)) * r;
	Point res1 = R * exp(Point(0, 1) * theta);
	Point res2 = R * exp(Point(0, 1) * (-theta));
	vector<Point> ret;
	ret.push_back(res1 + c1);
	ret.push_back(res2 + c1);
	return ret;
}

int toId(int i) {	//点solve.ps[i]が何番(>=0)の円に乗ってるか
	if (i == 0) return 0;
	if (i == 2 * n - 1) return n - 1;
	return (i - 1) / 2 + 1;
}

bool can_connect(int i, int j, Point s, Point e) {
	i = toId(i);
	j = toId(j);
	
	vector<Point> ps;
	for (int k = i; k <= j; k++) {
		double crs = cross(e - s, c[k] - s);
		if (crs * crs > r[k] * r[k] * norm(e - s)) return false;
		if (crs * crs == r[k] * r[k] * norm(e - s)) {
			if (k == j) return true;
			return false;
		}
		vector<Point> p = kouten_CL(c[k], r[k], s, e);
		ps.push_back(p[0]);
		ps.push_back(p[1]);
	}
	
	for (int k = 0; k + 3 < ps.size(); k += 2) {
		//誤差がなければ…、ギリギリ重なる場合はあって、ギリギリ重ならない場合はたぶん無い
		//ギリギリ重なる場合、誤差があるとギリギリ重ならないケースになるので、緩めにYes判定する.
		if (intersect_SS(ps[k], ps[k + 1], ps[k + 2], ps[k + 3]) == false) {
			return false;
		}
	}
	return true;
}

double solve() {
	int i, j;
	vector<Point> ps;
	vector<int> et[200];
	vector<double> ec[200];

	ps.push_back(c[0]);
	rep(i, n - 1) {
		Points p = kouten_CC(c[i], r[i], c[i + 1], r[i + 1]);
		ps.push_back(p[0]);
		ps.push_back(p[1]);
	}
	ps.push_back(c[n - 1]);
	
	/*rep(i, 2 * n) {
		cout << "ps[" << i << "] = (" << ps[i].real() << ", " << ps[i].imag() << ")" << endl;
	}*/
	
	rep(i, 2 * n) {
		for (j = i + 1; j < 2 * n; j++) {
			if (can_connect(i, j, ps[i], ps[j])) {
				double dist = abs(ps[j] - ps[i]);
				et[i].push_back(j);
				ec[i].push_back(dist);
				//cout << i << "->" << j << endl;
			}
		}
	}
	
	//DAGなので、ダイクストラは不要
	double dp[200];
	double INF = 1145141919;
	rep(i, 2 * n) dp[i] = INF;
	dp[0] = 0;
	rep(i, 2 * n) {
		rep(j, et[i].size()) {
			int nxt = et[i][j];
			dp[nxt] = min(dp[nxt], dp[i] + ec[i][j]);
		}
	}
	return dp[2 * n - 1];
}

signed main() {
	while (cin >> n) {
		if (!n) break;
		
		int i;
		rep(i, n) {
			double x, y;
			cin >> x >> y >> r[i];
			c[i] = Point(x, y);
		}
		double res = solve();
		printf("%.14Lf\n", res);
	}
	return 0;
}
