Problem: Consider a house plan. 

Let the house be represented by an infinite horizontal strip defined by the inequality  - h ≤ y ≤ h. Strictly outside the house there are two light sources at the points (0, f) and (0, - f). Windows are located in the walls, the windows are represented by segments on the lines y = h and y = - h. Also, the windows are arranged symmetrically about the line y = 0.

Your task is to find the area of the floor at the home, which will be lighted by the sources of light. 

<image>

Input

The first line of the input file contains three integers n, h and f (1 ≤ n ≤ 500, 1 ≤ h ≤ 10, h < f ≤ 1000). Next, n lines contain two integers each li, ri ( - 5000 ≤ li < ri ≤ 5000), each entry indicates two segments. Endpoints of the first segment are (li, h)-(ri, h), and endpoints of the second segment are (li, - h)-(ri, - h). These segments describe location of windows. Numbers in the lines are space-separated. It is guaranteed that no two distinct segments have common points. 

Output

Print the single real number — the area of the illuminated part of the floor with an absolute or relative error of no more than 10 - 4.

Examples

Input

1 1 2
-1 1


Output

10.0000000000


Input

2 2 4
-1 0
1 2


Output

23.3333333333

Note

The second sample test is shown on the figure. Green area is the desired area of the illuminated part of the floor. Violet segments indicate windows.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T alex(T x) {
  return (x > 0) ? x : -x;
}
template <class T>
void read(T& x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while ((ch < 48 || ch > 57) && ch != 45) ch = getchar();
  if (ch == 45) f = -1, ch = getchar();
  while (ch >= 48 && ch <= 57) x = x * 10 + ch - 48, ch = getchar();
  x *= f;
}
int sign(double x) {
  if (fabs(x) < 1e-9) return 0;
  if (x > 1e-9)
    return 1;
  else
    return -1;
}
struct Point {
  double x, y;
  Point() {}
  Point(double x, double y) : x(x), y(y) {}
  Point operator+(const Point& p) const { return Point(x + p.x, y + p.y); }
  Point operator-(const Point& p) const { return Point(x - p.x, y - p.y); }
  Point operator*(const double& d) const { return Point(x * d, y * d); }
  double operator^(const Point& p) const { return x * p.y - y * p.x; }
  double operator*(const Point& p) const { return x * p.x + y * p.y; }
  void print() { printf("%lf %lf\n", x, y); }
};
struct Line {
  Point s, v;
  double angle;
  Line() {}
  Line(Point s, Point v) : s(s), v(v) { angle = atan2(v.y, v.x); }
  bool operator<(const Line& l) const {
    return (sign(angle - l.angle) == 0) ? sign((l.s - s) ^ v) == 1
                                        : angle < l.angle;
  }
  void print() { printf("(%lf %lf)(%lf %lf)\n", s.x, s.y, v.x, v.y); }
};
int n, h, f;
int l[505], r[505];
double res;
Line que[505];
Point pnt[505];
Point inter(Line l1, Line l2) {
  double lambda = ((l2.s - l1.s) ^ l2.v) / (l1.v ^ l2.v);
  return l1.s + l1.v * lambda;
}
bool onleft(Point p, Line l) { return sign((p - l.s) ^ l.v) <= 0; }
vector<Point> HPI(vector<Line> l) {
  vector<Point> res;
  res.clear();
  sort(l.begin(), l.end());
  vector<Line> _l;
  _l.push_back(l[0]);
  for (int i = (1); i < (l.size()); i++)
    if (sign(l[i].angle - l[i - 1].angle) != 0) _l.push_back(l[i]);
  int head, tail;
  que[head = tail = 1] = _l[0];
  que[++tail] = _l[1];
  pnt[1] = inter(que[1], que[2]);
  for (int i = (2); i < (_l.size()); i++) {
    while (head < tail && !onleft(pnt[tail - 1], _l[i])) tail--;
    while (head < tail && !onleft(pnt[head], _l[i])) head++;
    pnt[tail] = inter(que[tail], _l[i]);
    que[++tail] = _l[i];
  }
  while (head < tail && !onleft(pnt[tail - 1], que[head])) tail--;
  while (head < tail && !onleft(pnt[head], que[tail])) head++;
  pnt[tail] = inter(que[head], que[tail]);
  for (int i = (head); i <= (tail); i++) res.push_back(pnt[i]);
  return res;
}
double getarea(vector<Point> p) {
  double res = 0.0;
  for (int i = (2); i < (p.size()); i++)
    res += (p[i] - p[0]) ^ (p[i] - p[i - 1]);
  return fabs(res) / 2.0;
}
int main() {
  read(n);
  read(h);
  read(f);
  for (int i = (1); i <= (n); i++) {
    read(l[i]), read(r[i]);
    res += 4.0 * (r[i] - l[i]) * f * h / (f - h);
  }
  for (int i = (1); i <= (n); i++)
    for (int j = (i); j <= (n); j++) {
      vector<Line> line;
      line.clear();
      Point a, b, c, d;
      a = Point(l[i], h);
      b = Point(1.0 * l[i] * (f + h) / (f - h), -h);
      c = Point(1.0 * r[i] * (f + h) / (f - h), -h);
      d = Point(r[i], h);
      line.push_back(Line(a, b - a));
      line.push_back(Line(b, c - b));
      line.push_back(Line(c, d - c));
      line.push_back(Line(d, a - d));
      a = Point(l[j], -h);
      b = Point(r[j], -h);
      c = Point(1.0 * r[j] * (f + h) / (f - h), h);
      d = Point(1.0 * l[j] * (f + h) / (f - h), h);
      line.push_back(Line(a, b - a));
      line.push_back(Line(b, c - b));
      line.push_back(Line(c, d - c));
      line.push_back(Line(d, a - d));
      double ans = getarea(HPI(line));
      if (i == j)
        res -= ans;
      else
        res -= 2 * ans;
    }
  printf("%lf\n", res);
}