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 <bits/stdc++.h>
using namespace std;

using ld = long double;

constexpr ld eps = 1e-8;

namespace geometry2d {


using point = std::complex<long double>;

bool eq(long double a, long double b) {
    return (std::abs(a-b) < eps);
}


long double dot(point a, point b) {
    return std::real(std::conj(a) * b);
}

long double cross(point a, point b) {
    return std::imag(std::conj(a) * b);
}


class segment {
public:
    segment()
        : a(point(0, 0)),
          b(point(0, 0))
    {}
    segment(point a_, point b_)
        : a(a_), b(b_)
    {}

    point a, b;
};

class line {
public:
    line()
        : a(point(0, 0)), b(point(0, 0))
    {}
    line(point a_, point b_)
        : a(a_), b(b_)
    {}
    line(segment s)
        : a(s.a), b(s.b)
    {}

    point a, b;
};

class circle {
public:
    circle()
        : p(point(0, 0)), r(0)
    {}
    circle(point p_, long double r_)
        : p(p_), r(r_)
    {}

    point p;
    long double r;
};


int ccw(point a, point b, point c) {
    b -= a; c -= a;
    if(cross(b, c) > eps)           return 1;  // a -> b -> c : counterclockwise
    if(cross(b, c) < -eps)          return -1; // a -> b -> c : clockwise
    if(dot(b, c) < 0)               return 2;  // c -> a -> b : line
    if(std::norm(b) < std::norm(c)) return -2; // a -> b -> c : line
    return 0;                                  // a -> c -> b : line
}


bool isis_ll(line l, line m) {
    return !eq(cross(l.b - l.a, m.b - m.a), 0);
}

bool isis_ls(line l, segment s) {
    return (cross(l.b - l.a, s.a - l.a) * cross(l.b - l.a, s.b - s.a) < eps);
}

bool isis_lp(line l, point p) {
    return (std::abs(cross(l.b - p, l.a - p)) < eps);
}

bool isis_sp(segment s, point p) {
    return (std::abs(s.a - p) + std::abs(s.b - p) - std::abs(s.b - s.a) < eps);
}

bool isis_ss(segment s, segment t) {
    return ccw(s.a, s.b, t.a) * ccw(s.a, s.b, t.b) <= 0
        && ccw(t.a, t.b, s.a) * ccw(t.a, t.b, s.b) <= 0;
}


point proj(line l, point p) {
    long double t = dot(p - l.a, l.a - l.b) / std::norm(l.a - l.b);
    return l.a + t * (l.a - l.b);
}

point is_ll(line s, line t) {
    point sv = s.b - s.a, tv = t.b - t.a;
    assert(cross(sv, tv) != 0);
    return s.a + sv * cross(tv, t.a - s.a) / cross(tv, sv);
}

long double dist_lp(line l, point p) {
    return std::abs(p - proj(l, p));
}

long double dist_ll(line l, line m) {
    return isis_ll(l, m) ? 0 : dist_lp(l, m.a);
}

long double dist_ls(line l, segment s) {
    return isis_ls(l, s) ? 0 : std::min(dist_lp(l, s.a), dist_lp(l, s.b));
}

long double dist_sp(segment s, point p) {
    point r = proj(line(s), p);
    return isis_sp(s, r) ? std::abs(r - p) : std::min(std::abs(s.a - p), std::abs(s.b - p));
}

long double dist_ss(segment s, segment t) {
    if(isis_ss(s, t)) return 0;
    long double d1 = std::min(dist_sp(s, t.a), dist_sp(s, t.b));
    long double d2 = std::min(dist_sp(t, s.a), dist_sp(t, s.b));
    return std::min(d1, d2);
}


std::vector<point> is_cc(circle const& c1, circle const& c2) {
    std::vector<point> res;
    long double d = std::abs(c1.p - c2.p);
    long double rc = (d * d + c1.r * c1.r - c2.r * c2.r) / (2 * d);
    long double dfr = c1.r * c1.r - rc * rc;
    if(std::abs(dfr) < eps) {
        dfr = 0.0;
    } else if(dfr < 0.0) {
        return res;
    }

    long double rs = std::sqrt(dfr);
    point diff = (c2.p - c1.p) / d;
    res.push_back(c1.p + diff * point(rc, rs));
    if(dfr != 0.0) {
        res.push_back(c1.p + diff * point(rc, -rs));
    }
    return res;
}


} // namespace geometry2d

using namespace geometry2d;

struct edge {
    int to;
    ld cost;
};

using edges = vector<edge>;
using graph = vector<edges>;

int main() {
    int n;
    while(cin >> n, n) {
        vector<circle> cs;
        vector<point> ps;
        for(int i=0; i<n; ++i) {
            ld x, y, r;
            cin >> x >> y >> r;
            cs.emplace_back(point(x, y), r);
        }
        ps.push_back(cs[0].p);
        for(int i=0; i<n-1; ++i) {
            auto cps = is_cc(cs[i], cs[i+1]);
            if(ccw(cs[i].p, cs[i+1].p, cps[0]) == -1) {
                swap(cps[0], cps[1]);
            }
            for(auto& cp : cps) {
                ps.push_back(move(cp));
            }
        }
        ps.push_back(cs[n-1].p);
        const int N = ps.size();
        graph g(N);
        for(int i=0; i<N; ++i) {
            for(int j=i+1; j<N; ++j) {
                bool ok = true;
                int idx1, idx2;
                if(i % 2 == 1) {
                    idx1 = i+2;
                } else {
                    idx1 = i+1;
                }
                if(j % 2 == 0) {
                    idx2 = j-1;
                } else {
                    idx2 = j;
                }
                for(int k=idx1; k<idx2; k+=2) {
                    if(ccw(ps[i], ps[j], ps[k]) == -1 || ccw(ps[i], ps[j], ps[k+1]) == 1) {
                        ok = false;
                        break;
                    }
                }
                if(ok) {
                    ld dist = abs(ps[i] - ps[j]);
                    g[i].push_back((edge){j, dist});
                    g[j].push_back((edge){i, dist});
                }
            }
        }
        vector<ld> d(N, 1e18);
        d[0] = 0;
        using P = pair<ld, int>;
        priority_queue<P, vector<P>, greater<P>> que;
        que.push(make_pair(0, 0));
        while(!que.empty()) {
            P p = que.top();
            que.pop();
            int v = p.second;
            if(d[v] < p.first) {
                continue;
            }
            for(auto& e : g[v]) {
                if(d[e.to] > d[v] + e.cost) {
                    d[e.to] = d[v] + e.cost;
                    que.push(make_pair(d[e.to], e.to));
                }
            }
        }
        cout << fixed << setprecision(10) << d[N-1] << endl;
    }
}
