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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stdio.h>
#include <queue>
#include <math.h>
#include <complex>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<VI> VVI;
#define rep(i,n) for (int i=0; i < (n); i++)

typedef double D;
typedef complex<D> P;
typedef pair<P, P> L;
typedef vector<P> VP;
const D EPS = 1e-9;
#define X real()
#define Y imag()
#define LE(n,m) ((n) < (m) + EPS)
#define GE(n,m) ((n) + EPS > (m))
#define EQ(n,m) (abs((n)-(m)) < EPS)

// 内積　dot(a,b) = |a||b|cosθ
D dot(P a, P b) {
  return (conj(a)*b).X;
}
// 外積　cross(a,b) = |a||b|sinθ
D cross(P a, P b) {
  return (conj(a)*b).Y;
}

// 点の進行方向
int ccw(P a, P b, P c) {
  b -= a;  c -= a;
  if (cross(b,c) >  EPS) return +1;  // counter clockwise
  if (cross(b,c) < -EPS) return -1;  // clockwise
  if (dot(b,c)   < -EPS) return +2;  // c--a--b on line
  if (norm(b) < norm(c)) return -2;  // a--b--c on line or a==b
  return 0;                          // a--c--b on line or a==c or b==c
}

/* 交差判定　直線・線分は縮退してはならない。接する場合は交差するとみなす。isecはintersectの略 */

// 直線と点
bool isecLP(P a1, P a2, P b) {
  return abs(ccw(a1, a2, b)) != 1;  // return EQ(cross(a2-a1, b-a1), 0); と等価
}

// 直線と直線
bool isecLL(P a1, P a2, P b1, P b2) {
  return !isecLP(a2-a1, b2-b1, 0) || isecLP(a1, b1, b2);
}

// 直線と線分
bool isecLS(P a1, P a2, P b1, P b2) {
  return cross(a2-a1, b1-a1) * cross(a2-a1, b2-a1) < EPS;
}

// 線分と線分
bool isecSS(P a1, P a2, P b1, P b2) {
  return ccw(a1, a2, b1)*ccw(a1, a2, b2) <= 0  &&  ccw(b1, b2, a1)*ccw(b1, b2, a2) <= 0;
}

// 線分と点
bool isecSP(P a1, P a2, P b) {
  return !ccw(a1, a2, b);
}


/* 距離　各直線・線分は縮退してはならない */

// 点pの直線aへの射影点を返す
P proj(P a1, P a2, P p) {
  return a1 + dot(a2-a1, p-a1)/norm(a2-a1) * (a2-a1);
}

// 点pの直線aへの反射点を返す
P reflection(P a1, P a2, P p) {
  return 2.0*proj(a1, a2, p) - p;
}

D distLP(P a1, P a2, P p) {
  return abs(proj(a1, a2, p) - p);
}

D distLL(P a1, P a2, P b1, P b2) {
  return isecLL(a1, a2, b1, b2) ? 0 : distLP(a1, a2, b1);
}

D distLS(P a1, P a2, P b1, P b2) {
  return isecLS(a1, a2, b1, b2) ? 0 : min(distLP(a1, a2, b1), distLP(a1, a2, b2));
}

D distSP(P a1, P a2, P p) {
  P r = proj(a1, a2, p);
  if (isecSP(a1, a2, r)) return abs(r-p);
  return min(abs(a1-p), abs(a2-p));
}

D distSS(P a1, P a2, P b1, P b2) {
  if (isecSS(a1, a2, b1, b2)) return 0;
  return min(min(distSP(a1, a2, b1), distSP(a1, a2, b2)), min(distSP(b1, b2, a1), distSP(b1, b2, a2)));
}

// 2直線の交点
P crosspointLL(P a1, P a2, P b1, P b2) {
  D d1 = cross(b2-b1, b1-a1);
  D d2 = cross(b2-b1, a2-a1);
  if (EQ(d1, 0) && EQ(d2, 0)) return a1;  // same line
  if (EQ(d2, 0)) throw "kouten ga nai";   // 交点がない
  return a1 + d1/d2 * (a2-a1);
}


/* 円 */

D distLC(P a1, P a2, P c, D r) {
  return max(distLP(a1, a2, c) - r, 0.0);
}

D distSC(P a1, P a2, P c, D r) {
  D dSqr1 = norm(c-a1), dSqr2 = norm(c-a2);
  if (dSqr1 < r*r ^ dSqr2 < r*r) return 0;  // 円が線分を包含するとき距離0ならここをORに変える
  if (dSqr1 < r*r & dSqr2 < r*r) return r - sqrt(max(dSqr1, dSqr2));
  return max(distSP(a1, a2, c) - r, 0.0);
}

VP crosspointLC(P a1, P a2, P c, D r) {
  VP ps;
  P ft = proj(a1, a2, c);
  if (!GE(r*r, norm(ft-c))) return ps;

  P dir = sqrt(max(r*r - norm(ft-c), 0.0)) / abs(a2-a1) * (a2-a1);
  ps.push_back(ft + dir);
  if (!EQ(r*r, norm(ft-c))) ps.push_back(ft - dir);
  return ps;
}

// 2円の交点
void crosspointCC(P a, D ar, P b, D br, VP& ps) {
  P ab = b-a;
  D d = abs(ab);
  D crL = (norm(ab) + ar*ar - br*br) / (2*d);
  if (EQ(d, 0) || ar < crL) return;

  P abN = ab * P(0, sqrt(ar*ar - crL*crL) / d);
  P cp = a + crL/d * ab;
  ps.push_back(cp + abN);
  if (!EQ(norm(abN), 0)) ps.push_back(cp - abN);
}

typedef vector<D> VD;
typedef vector<VD> VVD;
const D INF = 1e9;

int main() {
  int n;
  while(cin >> n, n) {
    VP cs(n);
    VI rs(n);
    rep(i,n) {
      int x, y;
      cin >> x >> y >> rs[i];
      cs[i] = P(x, y);
    }
    int m = 2*n;
    VVD graph(m, VD(m, INF));  // 0: src, 2i+1, 2i+2: i番目とi+1番目の円の交点, 2n-1:sink

    VP ps;
    ps.push_back(cs[0]);
    rep (i, n-1) {
      crosspointCC(cs[i], rs[i], cs[i+1], rs[i+1], ps);
    }
    ps.push_back(cs[n-1]);

    rep(i,m) for (int j = i+1; j < m; j++) {
      P a = ps[i];
      P b = ps[j];
      bool ok = true;
      for (int k = i+1; k < j && ok; k++) {
      	if (ccw(a,b, ps[k]) != (k % 2 == 0 ? -1 : 1)) ok = false;
      }
      if (ok) graph[i][j] = abs(a - b);
    }

    VD dist(m, INF);
    dist[0] = 0;
    rep (i,m) for (int j = i+1; j < m; j++) {
      dist[j] = min(dist[j], dist[i] + graph[i][j]);
    }
    // rep(i,m) {
    //   rep(j,m) cout << graph[i][j] << " ";
    //   cout << endl;
    // }
    printf("%.6lf\n", dist[m-1]);
  }

  return 0;
}