// O(n^2) dynamic-programming solution to slalom.
// David Garcia Soriano.
#include <algorithm>
#include <cassert>
#include <cstdio>
#include <cmath>
using namespace std;

const double eps = 1e-9, infinity = 1e300;

struct point {
  double x, y;
  point(double a = 0, double b = 0) : x(a), y(b) {}
};

double area2(point a, point b, point c) {
  return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}

// Returns true if c is to the left of ab
bool leftturn(point a, point b, point c) { return area2(a, b, c) > eps; }

bool norightturn(point a, point b, point c) { return area2(a, b, c) > -eps; }

double dist(const point& a, const point& b) { return hypot(a.x - b.x, a.y - b.y); }

point intersect(point p, point q, double y) {
  double dx = q.x - p.x, dy = q.y - p.y;
  return point(p.x + dx / dy * (y - p.y), y);
}

int main() {
  int n;
  while (scanf("%i", &n) == 1 && n != 0) {
    point interv[n + 1][2];
    double f[n + 1][2]; // f[i][s] = minimum distance from side s of interval i

    point p, e[2];

    scanf("%lf %lf\n", &p.x, &p.y);
    interv[0][0] = interv[0][1] = point(p.x, p.y);
    for (int i = 1; i <= n; ++i) {
      double y, x1, x2;
      scanf("%lf %lf %lf\n", &y, &x1, &x2);
      interv[i][0] = point(x1, y);
      interv[i][1] = point(x2, y);
    }

    f[n][0] = f[n][1] = 0;
    for (int i = n - 1; i >= 0; --i)
      for (int side = 0; side < 2; ++side) {
        point p(interv[i][side]);
        e[0] = interv[i + 1][0];
        e[1] = interv[i + 1][1];

        int j;
        f[i][side] = infinity;
        for (j = i + 1; j <= n; ++j) {
          if (leftturn(p, interv[j][1], e[0]) || leftturn(p, e[1], interv[j][0])) {
            // whole interval to the left or to the right
            break;
          } else {
            if (norightturn(p, e[0], interv[j][0])) {
              e[0] = interv[j][0];            // trim left endpoint
              f[i][side] = min(f[i][side], dist(p, interv[j][0]) + f[j][0]);
            }
            if (norightturn(p, interv[j][1], e[1])) {
              e[1] = interv[j][1];            // trim right endpoint
              f[i][side] = min(f[i][side], dist(p, interv[j][1]) + f[j][1]);
            }
          }
        }
        assert(j > i + 1);
        if (j > n) { // reached last interval
          double y = interv[n][0].y;
          e[0] = intersect(p, e[0], y);
          e[1] = intersect(p, e[1], y);

          double d;
          if (p.x > e[1].x) d = dist(p, e[1]);
          else if (p.x < e[0].x) d = dist(p, e[0]);
          else d = p.y - e[0].y;
          f[i][side] = min(f[i][side], d);
        }
        if (i == 0) break;
      }

    printf("%.11lf\n", f[0][0]);
  }
  return 0;
}