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 <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <functional>
#include <iostream>
#include <iomanip>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <utility>
#include <vector>

using namespace std;

typedef long long Long;
#define whole(xs) xs.begin(), xs.end()

// ヘロンの公式により三角形の面積を求める
double Heron(double a, double b, double c) {
    double s = (a + b + c) / 2;
    return sqrt(s * (s - a) * (s - b) * (s - c));
}

// 2次元座標上の点、あるいはベクトル
struct Point {
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) {}
    double abs() { return sqrt(x * x + y * y); }
    Point unit() {
        double l = abs();
        return Point(x / l, y / l);
    }
    Point operator+(const Point& p) const { return Point(x + p.x, y + p.y); }
    Point operator-() const { return Point(-x, -y); }
    Point operator-(const Point& p) const { return Point(x - p.x, y - p.y); }
    Point operator*(double a) const { return Point(a * x, a * y); }
    double operator*(const Point& p) const { return x * p.y + y * p.x; }
    Point operator/(double a) const { return Point(x / a, y / a); }
    Point R90() const { return Point(y, -x); }
    Point R180() const { return Point(-x, -y); }
    Point R270() const { return Point(-y, x); }
    static double Distance(const Point& a, const Point& b) {
        return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
    }
#define EPS (1e-9)
    static double CROSS(const Point& a, const Point& b) { return a.x * b.y - a.y * b.x; }
    static int CCW(Point a, Point b, Point c){
        b = b - a; c = c - a;
        if (CROSS(b, c) > EPS) return +1;  // 反時計周り
        if (CROSS(b, c) < -EPS) return -1; // 時計周り
        if (b * c < 0) return +2;          // c--a--b 直線
        if (b.abs() < c.abs()) return -2;  // a--b--c 直線
        return 0;                          // a--c--b 直線
    }
};
ostream& operator<<(ostream& os, const Point& p) {
    os << "(" << p.x << "," << p.y << ")";
    return os;
}

struct Segment {
    Point a, b;
    Segment(Point a, Point b) : a(a), b(b) {}
    bool DoesIntersect(const Segment& s) const {
        return Point::CCW(a, b, s.a) * Point::CCW(a, b, s.b) <= 0 &&
               Point::CCW(s.a, s.b, a) * Point::CCW(s.a, s.b, b) <= 0;
    }
};

struct Circle {
    Point o;
    double r;
    Circle(Point o, double r) : o(o), r(r) {}
    static pair<Point, Point> IPoints(const Circle& a, const Circle& b) {
        double d = Point::Distance(a.o, b.o);
        double h = Heron(a.r, b.r, d) / d * 2;
        double l = sqrt(a.r * a.r - h * h);
        Point e = (b.o - a.o).unit();
        Point proj = a.o + e * l;
        return make_pair(proj + e.R90() * h, proj + e.R270() * h);
    }
};

int N;
#define MN 100
double X[MN + 1]; 
double Y[MN + 1];
double R[MN + 1];
bool input() {
    cin >> N;
    if (N == 0) return false;
    for (int i = 0; i < N; i++) {
        cin >> X[i] >> Y[i] >> R[i];
    }
    return true;
}

void solve() {
    static const double INF = 1e20;
    static double G[MN * 2][MN * 2];
    vector<Point> Ps;
    Ps.push_back(Point(X[0], Y[0]));
    for (int i = 0; i < N - 1; i++) {
        Circle a(Point(X[i], Y[i]), R[i]),
               b(Point(X[i + 1], Y[i + 1]), R[i + 1]);
        pair<Point, Point> pp = Circle::IPoints(a, b);
        Ps.push_back(pp.first);
        Ps.push_back(pp.second);
    }
    Ps.push_back(Point(X[N - 1], Y[N - 1]));
    int N1 = 2 * N;
    for (int i = 0; i < N1; i++) for (int j = 0; j < N1; j++) {
        if (i == j) G[i][j] = 0;
        else G[i][j] = INF;
    }
    for (int i = 0; i < Ps.size(); i++) {
        for (int j = i + 1; j < Ps.size(); j++) {
            Point a = Ps[i];
            Point b = Ps[j];
            Segment s(a, b);
            int li = (i + 1) / 2 + 1,
                lj = (j + 1) / 2;
            for (int k = li; k < lj; k++) {
                if (!s.DoesIntersect(Segment(Ps[k * 2 - 1], Ps[k * 2]))) {
                    goto FAIL;
                }
            }
            G[i][j] = Point::Distance(a, b);
            FAIL:;
        }
    }
    for (int i = 0; i < N1; i++) {
        for (int j = i + 1; j < N1; j++) {
            G[0][j] = min(G[0][j], G[0][i] + G[i][j]);
        }
    }
    printf("%.8lf\n", G[0][2 * N - 1]);
}

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