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;

// conj(x) : complex conjugate,(0,1)->(0,-1)
// abs(x) : dist between(0,0) and x
// norm(x) : abs(x) * abs(x)
// arg(x) : argment,[-PI,PI]
using CP = complex<long double>;
#define X real()
#define Y imag()
const long double PI = acos(-1.0L);
const long double EPS = 1e-10;
// cos a
long double costh(long double a, long double b,
                  long double c) {
  return (b * b + c * c - a * a) / (2.0L * b * c);
}

// dot(a,b) = |a||b|cos x
long double dot(CP a, CP b) { return (conj(a) * b).X; }
// cross(a,b) : area of parallelogram
// sign : a-> b ,counter clockwise? + : -
long double cross(CP a, CP b) { return (conj(a) * b).Y; }
long double corner(CP a, CP b) {
  //[0,PI]
  return acos(dot(a, b) / (abs(a) * abs(b)));
}

CP projection(CP s, CP t, CP p) {
  CP base = t - s;
  long double r = dot(p - s, base) / norm(base);
  return s + base * r;
}

CP reflection(CP s, CP t, CP p) {
  CP tmp = (projection(s, t, p) - p);
  tmp *= 2;
  return p + tmp;
}

CP intersectionLL(CP a, CP b, CP c, CP d) {
  return a + (b - a) * (cross(d - c, c - a) /
                        cross(d - c, b - a));
}

bool on_seg(CP s, CP t, CP p) {
  // if not use end point, dot(s - p, t - p) < 0
  return abs(cross(s - p, t - p)) <= 1e-10 &&
         dot(s - p, t - p) <= 0;
}

// crossing lines? (a,b) and (c,d)
bool iscross(CP a, CP b, CP c, CP d) {
  // parallel
  if(abs(cross(a - b, c - d)) <= 1e-10) {
    return on_seg(a, b, c) || on_seg(a, b, d) ||
           on_seg(c, d, a) || on_seg(c, d, b);
  }
  CP isp = intersectionLL(a, b, c, d);
  return on_seg(a, b, isp) && on_seg(c, d, isp);
}

long double distLP(CP s, CP t, CP p) {
  return abs(cross(t - s, p - s) / abs(t - s));
}

// segmentver.
long double distSP(CP s, CP t, CP p) {
  if(dot(t - s, p - s) < 0) return abs(p - s);
  if(dot(s - t, p - t) < 0) return abs(p - t);
  return distLP(s, t, p);
}

// segment and segment
long double distSS(CP a, CP b, CP c, CP d) {
  long double res = 1e18;
  if(iscross(a, b, c, d)) return 0.0L;
  res = min(res, distSP(a, b, c));
  res = min(res, distSP(a, b, d));
  res = min(res, distSP(c, d, a));
  res = min(res, distSP(c, d, b));
  return res;
}

// counter clockwise
bool is_convex(const vector<CP> &v) {
  int n = v.size();
  for(int i = 0; i < n; ++i)
    if(cross(v[(i + 1) % n] - v[i],
             v[(i + 2) % n] - v[(i + 1) % n]) < -EPS)
      return 0;
  return 1;
}

vector<CP> convex_hull(vector<CP> &ps) {
  auto lmd = [&](const CP &l, const CP &r) {
    if(l.X != r.X) return l.X < r.X;
    return l.Y < r.Y;
  };
  vector<CP> qs;
  int psize = ps.size();
  sort(ps.begin(), ps.end(), lmd);
  int k = 0;
  qs.resize(psize * 2);
  for(int i = 0; i < psize; ++i) {
    while(k > 1 && cross(qs[k - 1] - qs[k - 2],
                         ps[i] - qs[k - 1]) <= 0)
      --k;
    qs[k++] = ps[i];
  }
  for(int i = psize - 2, t = k; i >= 0; --i) {
    while(k > t && cross(qs[k - 1] - qs[k - 2],
                         ps[i] - qs[k - 1]) <= 0)
      --k;
    qs[k++] = ps[i];
  }
  qs.resize(k - 1);
  return qs;
}

long double convex_diameter(vector<CP> &newv) {
  vector<CP> v = convex_hull(newv);
  int n = v.size(), i = 0, j = 0;
  if(n == 2) return abs(v[0] - v[1]);
  for(int k = 0; k < n; ++k) {
    if(v[k].X < v[i].X) i = k;
    if(v[k].X > v[j].X) j = k;
  }
  long double res = 0;
  int si = i, sj = j;
  while(i != sj || j != si) {
    res = max(res, abs(v[i] - v[j]));
    if(cross(v[(i + 1) % n] - v[i], v[(j + 1) % n] - v[j]) <
       0)
      (++i) %= n;
    else
      (++j) %= n;
  }
  return res;
}

vector<CP> convex_cut(const vector<CP> &v, const CP &s,
                      const CP &t) {
  vector<CP> res;
  int n = v.size();
  for(int i = 0; i < n; ++i) {
    CP nows = v[i], nowt = v[(i + 1) % n];
    if(cross(t - s, nows - s) >= -EPS) res.push_back(nows);
    if(cross(t - s, nows - s) * cross(t - s, nowt - s) < 0)
      res.push_back(intersectionLL(s, t, nows, nowt));
  }
  return res;
}

struct Circle {
  CP o;
  long double r;
  Circle(long double _x = 0.0L, long double _y = 0.0L,
         long double _r = 0.0L)
      : o(CP(_x, _y)), r(_r) {}
  Circle(CP _o, long double _r = 0.0) : o(_o), r(_r) {}
};

void intersectionCC(Circle c1, Circle c2, CP &res1,
                    CP &res2) {
  long double d = abs(c1.o - c2.o);
  long double arg = acos(costh(c2.r, c1.r, d));
  res1 = (c2.o - c1.o) * CP(cos(arg), sin(arg));
  res1 *= c1.r / abs(res1);
  res1 += c1.o;
  res2 = reflection(c1.o, c2.o, res1);
  if(res1.X > res2.X ||
     (res1.X == res2.X && res1.Y > res2.Y))
    swap(res1, res2);
}

void intersectionCL(Circle ci, CP s, CP t, CP &res1,
                    CP &res2) {
  res1 = res2 = projection(s, t, ci.o);
  long double r = sqrtl(ci.r * ci.r - norm(res1 - ci.o));
  t -= s;
  t *= r / abs(t);
  res1 += t;
  res2 -= t;
  if(res1.X > res2.X ||
     (res1.X == res2.X && res1.Y > res2.Y))
    swap(res1, res2);
}

void contactCP(Circle ci, CP p, CP &res1, CP &res2) {
  long double d = abs(ci.o - p);
  if(abs(d - ci.r) <= EPS) {
    res1 = res2 = p;
    return;
  }
  else if(d < ci.r) {
    res1 = res2 = CP(-1e18, -1e18);
    return;
  }
  long double arg = asin(ci.r / d);
  res1 = (ci.o - p) * CP(cos(arg), sin(arg));
  res1 *= (d * cos(arg)) / abs(res1);
  res1 += p;
  res2 = reflection(p, ci.o, res1);
  if(res1.X > res2.X ||
     (res1.X == res2.X && res1.Y > res2.Y))
    swap(res1, res2);
}

long double polygonarea(const vector<CP> &v) {
  int n = v.size();
  long double res = 0;
  for(int i = 0; i < n; ++i)
    res +=
        (v[(i - 1 + n) % n].X - v[(i + 1) % n].X) * v[i].Y;
  return res / 2.0L;
}

long long calc_clockwise(CP s, CP t, CP p) {
  t -= s;
  p -= s;
  if(cross(t, p) > EPS) return 1;   // "COUNTER_CLOCKWISE"
  if(cross(t, p) < -EPS) return -1; //"CLOCK_WISE"
  if(dot(t, p) < 0) return 2;       // ONLINE_BACK
  if(norm(t) < norm(p)) return -2;  // ONLINE_FRONT
  return 0;                         // ON_SEGMENT
}

long long n;
vector<vector<long double>> dist;
vector<CP> points;
vector<int> id;
vector<Circle> circles;

long double solve();

int main() {
  cout << fixed << setprecision(10);
  while(1) {
    cin >> n;
    if(n == 0) break;
    circles.resize(n);
    for(int i = 0; i < n; ++i) {
      int x, y;
      cin >> x >> y >> circles[i].r;
      circles[i].o = CP(x, y);
    }
    cout << solve() << endl;
  }
  return 0;
}

long double solve() {
  points.clear();
  dist.clear();
  id.clear();
  points.push_back(circles[0].o);
  id.push_back(0);
  for(int i = 0; i < n - 1; ++i) {
    points.push_back(CP(0, 0));
    points.push_back(CP(0, 0));
    id.push_back(i + 1);
    id.push_back(i + 1);
    intersectionCC(circles[i], circles[i + 1],
                   points[1 + 2 * i], points[2 + 2 * i]);
  }
  points.push_back(circles[n - 1].o);
  id.push_back(n);
  int psize = points.size();
  dist.assign(psize, vector<long double>(psize, 1e18));
  for(int i = 0; i < psize; ++i)
    for(int j = i; j < psize; ++j) {
      bool ch = 1;
      CP nowp = points[i], st = points[i], go = points[j];
      if(i == 0) {
        CP dummy1, dummy2;
        intersectionCL(circles[0], st, go, dummy1, dummy2);
        if(norm(dummy2 - go) < norm(dummy1 - go))
          dummy1 = dummy2;
        st = dummy1;
      }
      if(j == psize - 1) {
        CP dummy1, dummy2;
        intersectionCL(circles[n - 1], st, go, dummy1,
                       dummy2);
        if(norm(dummy2 - st) < norm(dummy1 - st))
          dummy1 = dummy2;
        go = dummy1;
      }
      nowp = st;
      for(int k = max(id[i], 1); k < id[j]; ++k) {
        if(distLP(st, go, circles[k].o) >
               circles[k].r + EPS ||
           abs(nowp - circles[k].o) > circles[k].r + EPS) {
          ch = 0;
          break;
        }
        CP p1(0, 0), p2(0, 0);
        intersectionCL(circles[k], st, go, p1, p2);
        if(norm(p2 - points[i]) < norm(p1 - points[i]))
          swap(p1, p2);
        nowp = p2;
      }
      if(ch) dist[i][j] = abs(points[i] - points[j]);
    }
  for(int k = 0; k < psize; ++k)
    for(int i = 0; i < psize; ++i)
      for(int j = 0; j < psize; ++j)
        dist[i][j] =
            min(dist[i][j], dist[i][k] + dist[k][j]);
  return dist[0][psize - 1];
}

