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
#define _USE_MATH_DEFINES
#include "bits/stdc++.h"
#define FOR(i, a, b) for (int i = a; i < b; i++)
#define REP(i, n) FOR(i, 0, n)
#define RFOR(i, a, b) for (int i = b - 1; i >= a; i--)
#define RREP(i, n) RFOR(i, 0, n)
#define rep(i, a, b) for (auto i = a; i < b; i++)
#define rrep(i, a, b) for (auto i = a; i > b; i--)
#define range(i, a, b) (a <= i && i < b)
#define fi first
#define fs first
#define se second
#define sc second
//#define int long long
using namespace std;
using vec = vector<int>;
using mat = vector<vec>;
#define ll long long

#define mp make_pair
#define all(v) begin(v), end(v)
//#define M_PI 3.14159265353238
#define pb push_back

namespace geo
{
#define IL inline
	typedef double D;
	typedef bool B;
	typedef complex<D> P;
#define X real()
#define Y imag()
	ostream &operator<<(ostream &out, const P &point1) // Pの出力
	{
		out << '(' << point1.X << ", " << point1.Y << ')';
		return out;
	}
	B comp(const P &l, const P &r)
	{
		return (l.X == r.X) ? l.Y < r.Y : l.X < r.X;
	}
	typedef pair<P, P> L;  //line
	typedef pair<P, P> LS; //line segment
	typedef pair<P, D> C;  //circle
	typedef vector<P> Poly;
	const D EPS = 1e-8;
	//Decompotision Macro
#define DCl(a, b, l) \
  P a, b;            \
  tie(a, b) = l
#define DCc(a, b, c) \
  P a;               \
  D b;               \
  tie(a, b) = c

//A dot B
	IL D dot(P a, P b) { return a.X * b.X + a.Y * b.Y; }
	//A cross B
	IL D cross(P a, P b) { return a.X * b.Y - a.Y * b.X; }
	IL D ccw(P a, P b, P c)
	{
		return cross(b - a, c - a);
	}

	// 未検証!!
	IL P rotation(P a, double angle) {
		a *= exp(complex<double>(.0, angle * M_PI / 180.0));
		return a;
	}

	IL int sgn(P a, P b, P c)
	{
		if (cross(b - a, c - a) > EPS)
			return 1;
		if (cross(b - a, c - a) < -EPS)
			return -1;
		if (dot(b - a, c - a) < -EPS)
			return 2; // b-a-c
		if (dot(a - b, c - b) < -EPS)
			return -2; // a-b-c
		return 0;
	}
	IL D signed_distance_P_L(P p, L l)
	{
		DCl(a, b, l);
		return ccw(a, b, p) / abs(b - a);
	}

	IL D distance_P_L(P p, L l) // 点と直線の距離
	{
		return abs(signed_distance_P_L(p, l));
	}

	IL D distance_P_P(P p1, P p2) // 二点間距離
	{
		return abs(p1 - p2);
	}

	IL D distance_P_LS(P p, LS ls) // 点と線分の距離
	{
		DCl(a, b, ls);
		if (dot(b - a, p - a) < EPS)
			return abs(p - a);
		if (dot(a - b, p - b) < EPS)
			return abs(p - b);
		return abs(cross(b - a, p - a)) / abs(b - a);
	}

	IL B is_L_L(L p, L q)
	{
		DCl(a, b, p);
		DCl(c, d, q);
		return abs(cross(a - b, c - d)) > EPS ||
			abs(cross(a - b, d - b)) < EPS;
	}
	IL B is_L_LS(L l, LS ls)
	{
		DCl(a, b, l);
		DCl(c, d, ls);
		return ccw(a, b, c) * ccw(a, b, d) < EPS;
	}
	IL B is_LS_LS(LS p, LS q)
	{
		DCl(a, b, p);
		DCl(c, d, q);
		return sgn(a, b, c) * sgn(a, b, d) <= 0 && sgn(c, d, a) * sgn(c, d, b) <= 0;
	}

	IL D distance_LS_LS(LS p, LS q)
	{
		if (is_LS_LS(p, q))
			return 0;
		return min({ distance_P_LS(p.fi, q),
					distance_P_LS(p.se, q),
					distance_P_LS(q.fi, p),
					distance_P_LS(q.se, p) });
	}

	//caution: have to use isintersected_L_L before this function
	IL P cross_L_L(L p, L q)
	{
		DCl(a, b, p);
		D d1 = signed_distance_P_L(a, q);
		D d2 = signed_distance_P_L(b, q);
		return (a * d2 - b * d1) / (d2 - d1);
	}

	IL B is_C_C(C p, C q)
	{
		DCc(cp, rp, p);
		DCc(cq, rq, q);
		D d = abs(cp - cq);
		return abs(rp - rq) - d < EPS &&
			rp + rq - d > -EPS;
	}
	IL LS cross_C_C(C p, C q)
	{
		DCc(cp, rp, p);
		DCc(cq, rq, q);
		D d = abs(cp - cq);
		D rc = (d * d + rp * rp - rq * rq) / (2 * d);
		D rs = sqrt(rp * rp - rc * rc);
		P diff = (cq - cp) / d;
		return { cp + P(rc, rs) * diff, cp + P(rc, -rs) * diff };
	}
	IL Poly convex_hull(Poly p)
	{
		int n = p.size();
		Poly res(2 * n);
		int k = 0, t;
		sort(all(p), comp);
		rep(i, 0, n)
		{
			while (k > 1 && ccw(res[k - 2], res[k - 1], p[i]) < EPS)
				k--;
			res[k++] = p[i];
		}
		t = k;
		rep(j, 0, n - 1)
		{
			int i = n - 2 - j;
			while (k > t && ccw(res[k - 2], res[k - 1], p[i]) < EPS)
				k--;
			res[k++] = p[i];
		}
		res.resize(k - 1);
		return res;
	}
	IL B convex_in(Poly &l, P &p)
	{
		int a = 0, b = l.size(), c;
		D A, C;
		P g = (l[a] + l[b - 1] + l[b / 2]) / 3.0;
		while (b - a > 1)
		{
			c = (a + b) / 2;
			A = cross(l[a] - g, p - l[a]);
			C = cross(l[c] - g, p - l[c]);
			if (cross(l[a] - g, l[c] - g) >= 0)
			{
				if (A > -EPS && C < -EPS)
					b = c;
				else
					a = c;
			}
			else
			{
				if (C < -EPS || A > -EPS)
					b = c;
				else
					a = c;
			}
		}
		return (cross(l[b % l.size()] - l[a], p - l[b % l.size()]) > -EPS);
	}

	//not necessary?
	IL D convex_area(Poly &l)
	{
		double res = 0;
		int n = l.size();
		rep(i, 2, n)
			res += (abs(cross(l[i - 1] - l[0], l[i] - l[i - 1]))) / 2.0;
		return res;
	}
	IL D area(Poly &l)
	{
		double res = 0;
		int n = l.size();
		rep(i, 0, n)
			res += cross(l[i], l[(i + 1) % n]);
		return abs(res) / 2.0;
	}
	inline vector<L> tangent_line(C c, P p)
	{
		vector<L> res;
		DCc(cp, r, c);
		P v = p - cp;
		D dist = abs(v);
		if (dist - r > -EPS)
		{
			double _cos = r / dist;
			double _sin = sqrt(1.0 - _cos * _cos);
			v = v * (r / dist);
			{
				P e = P(_cos * v.X - _sin * v.Y, _sin * v.X + _cos * v.Y);
				res.push_back({ p, cp + e });
			}
			if (dist - r > EPS)
			{
				P e = P(_cos * v.X + _sin * v.Y, -_sin * v.X + _cos * v.Y);
				res.push_back({ p, cp + e });
			}
		}
		return res;
	}
	IL vector<L> inner_common_tangent_line(C p, C q)
	{
		vector<L> res;
		DCc(Mp, Rp, p);
		DCc(Mq, Rq, q);
		D dist = abs(Mp - Mq);
		if (dist - (Rp + Rq) > -EPS)
		{
			P mid = (Mq * Rp + Mp * Rq) / (Rp + Rq);
			res = tangent_line(p, mid);
		}
		return res;
	}
	inline vector<L> outer_common_tangent_line(C p, C q)
	{
		vector<L> res;
		DCc(Mp, Rp, p);
		DCc(Mq, Rq, q);
		P v = Mp - Mq;
		D dist = abs(v);
		if (dist - abs(Rp - Rq) > -EPS)
		{
			if (abs(Rp - Rq) < EPS)
			{
				P e = v * P(0, 1);
				e = e / dist;
				res.pb(L(Mp + e, Mq + e));
				res.pb(L(Mp - e, Mq - e));
			}
			else
			{
				P mid = Mp + (Mq - Mp) * Rp / (Rp - Rq);
				res = tangent_line(p, mid);
			}
		}
		return res;
	}

	vector<Poly> decomposition(Poly p)
	{
		int n = p.size();
		vector<Poly> res;
		while (n >= 3)
		{
			[&]() {
				rep(i, 0, n)
				{
					P a = p[i];
					P b = p[(i + 1) % n];
					P c = p[(i + 2) % n];
					if (ccw(a, b, c) < 0)
						continue;
					if ([&]() {
						rep(j, 3, n)
						{
							P d = p[(i + j) % n];
								if (ccw(a, b, d) > -EPS && ccw(b, c, d) > -EPS && ccw(c, a, d) > -EPS)
									return true;
						}
						return false;
					}())
						continue;
						Poly nxt;
						rep(j, 0, n) if ((i + j) % n != (i + 1) % n)
							nxt.push_back(p[(i + j) % n]);
						swap(p, nxt);
						res.push_back({ a, b, c });
						n--;
						return;
				}
			}();
		}
		return res;
	}
	int in_triangle(P p, Poly q)
	{
		rep(i, 0, 3) if (abs(p - q[i]) < EPS) return 3; //on point
		rep(i, 0, 3)
		{
			int s = sgn(q[i], q[(i + 1) % 3], p);
			if (s != 1)
			{
				if (s == 0)
					return 2; //on line
				else
					return 0; // not in
			}
		}
		return 1; // inside triangle
	}
	int in_C(P q, C c) // 円での点の内外判定(円周上も判定)
	{
		DCc(p, r, c);
		if (abs(r - abs(p - q)) < EPS)
			return 2; // 円周
		return r - abs(p - q) > -EPS ? 1 : 0;
	}
	int is_triangle_C(Poly t, C c)
	{
		t = convex_hull(t);
		DCc(p, r, c);
		int cnt = 0;
		rep(i, 0, 3)
			cnt += (in_C(t[i], c) > 0);
		if (cnt == 3)
			return 3; // t in c
		if (cnt == 0)
		{
			double d = distance_P_LS(p, LS(t[2], t[0]));
			rep(i, 0, 2)
				d = min(d, distance_P_LS(p, LS(t[i], t[i + 1])));
			if (r - d > EPS)
				return 1; //intersect
			if (in_triangle(p, t))
				return 2; //c in t
			else if (r - d < -EPS)
				return 0; //not intersect
		}
		return 1; //intersect
	}

	//p must be convex and ccw
	LS diameter(Poly &p)
	{
		int n = p.size();
		int is = 0, js = 0;
		rep(i, 0, n) if (p[i].Y > p[is].Y)
			is = i;
		rep(i, 0, n) if (p[i].Y < p[js].Y)
			js = i;
		LS res = { p[is], p[js] };
		{
			int i = is;
			int j = js;
			do
			{
				int ni = (i + 1) % n;
				int nj = (j + 1) % n;
				if (cross(p[ni] - p[i], p[nj] - p[j]) < EPS)
					i = ni;
				else
					j = nj;
				if (abs(res.fi - res.se) < abs(p[i] - p[j]))
					res = LS(p[i], p[j]);
			} while (i != is || j != js);
		}
		return res;
	}

	Poly cross_C_L(C c, L v)
	{
		DCc(p, r, c);
		DCl(a, b, v);
		a -= p;
		b -= p;
		P u = a - b;
		u = P(-u.Y, u.X);
		P x = cross_L_L({ a, b }, { P(0, 0), u });
		D d = abs(x);
		Poly res;
		if (d - r > EPS)
			return res;
		D y = sqrt(r * r - d * d);
		res.pb(p + x + (b - a) * y / abs(b - a));
		//if(d-r>-EPS)return res;
		res.pb(p + x + (a - b) * y / abs(b - a));
		return res;
	}

	pair<C, C> GetC_p_p_r(P p1, P p2, D r) // 二点と半径から２つの円を求める
	{
		D d = distance_P_P(p1, p2);
		assert(d <= 2. * r); // 二点間の距離は2 * r以内じゃないと同じ円周上に存在することはできない.
		P mid = (p1 + p2) / 2.;
		P NormalVector = (p1 - p2) / d * sqrt(r * r - d * d / 4.);
		NormalVector = P(NormalVector.Y, -NormalVector.X);
		P center1 = mid + NormalVector;
		P center2 = mid - NormalVector;
		return mp(C(center1, r), C(center2, r));
	}
}

double dis[300][300];

void solve(int &n) {
	cout.precision(12); cout << fixed;
	using namespace geo;
	rep(i, 0, 300) rep(j, 0, 300) dis[i][j] = 1000000000.;
	rep(i, 0, 300) dis[i][i] = 0.;
	vector<C> circles;
	rep(i, 0, n) {
		int x, y, r; cin >> x >> y >> r;
		circles.pb(pair<P, D>(P(x, y), r));
	}
	vector<LS> lss;
	vector<P> ps;
	ps.push_back(circles[0].first);
	int now = 1;
	rep(i, 0, n - 1) {
		LS ls = cross_C_C(circles[i], circles[i + 1]);
		P now1 = ls.first;
		P now2 = ls.second;
		rep(j, 0, ps.size()) {
			bool ok = true;
			rep(k, (j + 1) / 2, lss.size()) {
				if (!is_LS_LS(LS(now1, ps[j]), LS(lss[k]))) ok = false;
			}
			if (ok) {
				dis[j][now] = min(dis[j][now], distance_P_P(now1, ps[j]));
				dis[now][j] = min(dis[now][j], distance_P_P(now1, ps[j]));
			}
		}
		rep(j, 0, ps.size()) {
			bool ok = true;
			rep(k, (j + 1) / 2, lss.size()) {
				if (!is_LS_LS(LS(now2, ps[j]), LS(lss[k]))) ok = false;
			}
			if (ok) {
				dis[j][now + 1] = min(dis[j][now + 1], distance_P_P(now2, ps[j]));
				dis[now + 1][j] = min(dis[now + 1][j], distance_P_P(now2, ps[j]));
			}
		}
		ps.pb(now1);
		ps.pb(now2);
		lss.pb(ls);
		now += 2;
	}
	{
		P now1 = circles.back().first;
		rep(j, 0, ps.size()) {
			bool ok = true;
			rep(k, (j + 1) / 2, lss.size()) {
				if (!is_LS_LS(LS(now1, ps[j]), LS(lss[k]))) ok = false;
			}
			if (ok) {
				dis[j][now] = min(dis[j][now], distance_P_P(now1, ps[j]));
				dis[now][j] = min(dis[now][j], distance_P_P(now1, ps[j]));
			}
		}
	}
	rep(k, 0, 300) rep(i, 0, 300) rep(j, 0, 300) {
		dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
	}
	cout << dis[0][now] << endl;
}

int main() {
	int n;
	while (cin >> n, n) solve(n);
}
