Problem: Mikhail the Freelancer dreams of two things: to become a cool programmer and to buy a flat in Moscow. To become a cool programmer, he needs at least p experience points, and a desired flat in Moscow costs q dollars. Mikhail is determined to follow his dreams and registered at a freelance site.

He has suggestions to work on n distinct projects. Mikhail has already evaluated that the participation in the i-th project will increase his experience by ai per day and bring bi dollars per day. As freelance work implies flexible working hours, Mikhail is free to stop working on one project at any time and start working on another project. Doing so, he receives the respective share of experience and money. Mikhail is only trying to become a cool programmer, so he is able to work only on one project at any moment of time.

Find the real value, equal to the minimum number of days Mikhail needs to make his dream come true.

For example, suppose Mikhail is suggested to work on three projects and a1 = 6, b1 = 2, a2 = 1, b2 = 3, a3 = 2, b3 = 6. Also, p = 20 and q = 20. In order to achieve his aims Mikhail has to work for 2.5 days on both first and third projects. Indeed, a1·2.5 + a2·0 + a3·2.5 = 6·2.5 + 1·0 + 2·2.5 = 20 and b1·2.5 + b2·0 + b3·2.5 = 2·2.5 + 3·0 + 6·2.5 = 20.

Input

The first line of the input contains three integers n, p and q (1 ≤ n ≤ 100 000, 1 ≤ p, q ≤ 1 000 000) — the number of projects and the required number of experience and money.

Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 1 000 000) — the daily increase in experience and daily income for working on the i-th project.

Output

Print a real value — the minimum number of days Mikhail needs to get the required amount of experience and money. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6. 

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.

Examples

Input

3 20 20
6 2
1 3
2 6


Output

5.000000000000000


Input

4 1 1
2 3
3 2
2 3
3 2


Output

0.400000000000000

Note

First sample corresponds to the example in the problem statement.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const signed long long INF = 1000000100;
const long double EPS = 1e-9;
const long double Pi = 2 * atanl(1.0);
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &V) {
  os << "[";
  for (int(i) = (0); (i) < (int(V.size())); (i)++)
    os << V[i] << ((i == int(V.size()) - 1) ? "" : ",");
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &S) {
  os << "(";
  for (T i : S) os << i << (i == *S.rbegin() ? "" : ",");
  return os << ")";
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const map<T, U> &M) {
  os << "{";
  for (pair<T, U> i : M) os << i << (i.X == M.rbegin()->X ? "" : ",");
  return os << "}";
}
inline bool is_zero(const long double &first) {
  return first >= -EPS and first <= EPS;
}
inline int sign(const long double &first) {
  if (first < -EPS)
    return -1;
  else if (first > EPS)
    return 1;
  else
    return 0;
}
inline pair<long double, long double> operator+(
    const pair<long double, long double> &a,
    const pair<long double, long double> &b) {
  return pair<long double, long double>(a.first + b.first, a.second + b.second);
}
inline pair<long double, long double> operator-(
    const pair<long double, long double> &a,
    const pair<long double, long double> &b) {
  return pair<long double, long double>(a.first - b.first, a.second - b.second);
}
inline pair<long double, long double> operator*(
    const pair<long double, long double> &a, long double t) {
  return pair<long double, long double>(a.first * t, a.second * t);
}
inline pair<long double, long double> operator/(
    const pair<long double, long double> &a, long double t) {
  return pair<long double, long double>(a.first / t, a.second / t);
}
inline bool operator==(const pair<long double, long double> &a,
                       const pair<long double, long double> &b) {
  return is_zero(a.first - b.first) and is_zero(a.second - b.second);
}
inline bool operator!=(const pair<long double, long double> &a,
                       const pair<long double, long double> &b) {
  return !is_zero(a.first - b.first) or !is_zero(a.second - b.second);
}
pair<long double, long double> FAIL = make_pair(1e30, 1e30);
inline long double cp(const pair<long double, long double> &a,
                      const pair<long double, long double> &b) {
  return a.first * b.second - a.second * b.first;
}
inline long double dp(const pair<long double, long double> &a,
                      const pair<long double, long double> &b) {
  return a.first * b.first + a.second * b.second;
}
inline long double pieprzu(const pair<long double, long double> &a,
                           const pair<long double, long double> &b) {
  return sqrtl((b.first - a.first) * (b.first - a.first) +
               (b.second - a.second) * (b.second - a.second));
}
inline long double distance_point_line(
    const pair<long double, long double> &p,
    const pair<pair<long double, long double>, pair<long double, long double> >
        &u) {
  return cp(p - u.first, p - u.second) / pieprzu(u.first, u.second);
}
inline bool on_right(const pair<long double, long double> &a,
                     const pair<long double, long double> &b,
                     const pair<long double, long double> &c) {
  return sign(cp(c - a, b - a)) == 1;
}
inline bool on_left(const pair<long double, long double> &a,
                    const pair<long double, long double> &b,
                    const pair<long double, long double> &c) {
  return sign(cp(c - a, b - a)) == -1;
}
inline double polygon_area(const vector<pair<long double, long double> > &A) {
  long double result = 0.0;
  if (int(A.size()) <= 2) return 0;
  for (int(i) = (2); (i) < (int(A.size())); (i)++)
    result += cp(A[i] - A[0], A[i - 1] - A[0]);
  return abs(result / 2);
}
inline pair<long double, long double> projection(
    const pair<long double, long double> &p,
    const pair<pair<long double, long double>, pair<long double, long double> >
        &u) {
  return u.first +
         ((u.second - u.first) * (dp(p - u.first, u.second - u.first) /
                                  dp(u.second - u.first, u.second - u.first)));
}
inline bool between(const pair<long double, long double> &a,
                    const pair<long double, long double> &b,
                    const pair<long double, long double> &c) {
  return sign(min(b.first, c.first) - a.first) <= 0 and
         sign(a.first - max(b.first, c.first)) <= 0 and
         sign(min(b.second, c.second) - a.second) <= 0 and
         sign(a.second - max(b.second, c.second)) <= 0;
}
inline bool on_segment(const pair<long double, long double> &p,
                       const pair<pair<long double, long double>,
                                  pair<long double, long double> > &u) {
  return between(p, u.first, u.second) and pieprzu(p, projection(p, u)) < EPS;
}
inline long double distance_segment_point(
    const pair<pair<long double, long double>, pair<long double, long double> >
        &u,
    const pair<long double, long double> &p) {
  if (on_segment(projection(p, u), u))
    return fabs(distance_point_line(p, u));
  else
    return min(pieprzu(u.first, p), pieprzu(u.second, p));
}
inline long double angle(const pair<long double, long double> &a,
                         const pair<long double, long double> &o,
                         const pair<long double, long double> &b) {
  long double result = atan2l(a.first - o.first, a.second - o.second) -
                       atan2l(b.first - o.first, b.second - o.second);
  if (sign(result) == -1) result += 2 * Pi;
  return result;
}
inline pair<long double, long double> point_reflection(
    const pair<long double, long double> &s,
    const pair<long double, long double> &p) {
  return p + (s - p) * 2;
}
inline pair<long double, long double> axial_reflection(
    const pair<long double, long double> &p,
    const pair<pair<long double, long double>, pair<long double, long double> >
        &u) {
  return point_reflection(p, projection(p, u));
}
inline int triangle_inequality(long double &p, long double q, long double r) {
  int a = sign(p + q - r);
  int b = sign(p + r - q);
  int c = sign(q + r - p);
  if (a == -1 or b == -1 or c == -1) return -1;
  if (a == 0 or b == 0 or c == 0) return 0;
  return 1;
}
vector<pair<long double, long double> > convex_hull(
    vector<pair<long double, long double> > A) {
  if (int(A.size()) <= 1) return A;
  sort(A.begin(), A.end());
  vector<pair<long double, long double> > R{A[0], A[1]};
  for (int(i) = (2); (i) < (int(A.size())); (i)++) {
    while (int(R.size()) >= 2 and
           !on_right(R[R.size() - 2], R[R.size() - 1], A[i]))
      R.pop_back();
    R.push_back(A[i]);
  }
  int u = int(A.size());
  for (int(i) = (int(A.size()) - 2); (i) >= (0); (i)--) {
    while (int(R.size()) > u and
           !on_right(R[R.size() - 2], R[R.size() - 1], A[i]))
      R.pop_back();
    R.push_back(A[i]);
  }
  R.pop_back();
  return move(R);
}
inline pair<long double, long double> rotate90_origin(
    const pair<long double, long double> &p) {
  return pair<long double, long double>(p.second, -p.first);
}
inline pair<long double, long double> rotate90(
    const pair<long double, long double> &p,
    const pair<long double, long double> &o) {
  return o + rotate90_origin(p - o);
}
inline pair<long double, long double> rotate_origin(
    const pair<long double, long double> &p, long double theta) {
  theta = -theta;
  long double s = sinl(theta);
  long double c = cosl(theta);
  return pair<long double, long double>(p.first * c - p.second * s,
                                        p.first * s + p.second * c);
}
inline pair<long double, long double> rotate(
    const pair<long double, long double> &p,
    const pair<long double, long double> &o, long double theta) {
  return o + rotate_origin(p - o, theta);
}
inline bool parallel(const pair<pair<long double, long double>,
                                pair<long double, long double> > &u,
                     const pair<pair<long double, long double>,
                                pair<long double, long double> > &v) {
  return is_zero(cp(u.second - u.first, v.second - v.first));
}
bool perpendicular(const pair<pair<long double, long double>,
                              pair<long double, long double> > &u,
                   const pair<pair<long double, long double>,
                              pair<long double, long double> > &v) {
  return is_zero(dp(u.second - u.first, v.second - v.first));
}
inline pair<pair<long double, long double>, pair<long double, long double> >
segment_bisector(const pair<pair<long double, long double>,
                            pair<long double, long double> > &u) {
  pair<long double, long double> p = (u.first + u.second) / 2;
  return pair<pair<long double, long double>, pair<long double, long double> >(
      p, rotate90(u.second, p));
}
inline pair<pair<long double, long double>, pair<long double, long double> >
angle_bisector(const pair<long double, long double> &a,
               const pair<long double, long double> &o,
               const pair<long double, long double> &b) {
  return pair<pair<long double, long double>, pair<long double, long double> >(
      o, rotate(b, o, angle(a, o, b) / 2));
}
inline pair<long double, long double> intersection_line_line(
    const pair<pair<long double, long double>, pair<long double, long double> >
        &u,
    const pair<pair<long double, long double>, pair<long double, long double> >
        &v) {
  long double p = cp(u.second - v.first, u.second - u.first);
  long double q = cp(v.second - v.first, u.second - u.first);
  if (is_zero(q)) return FAIL;
  return v.first + (v.second - v.first) * (p / q);
}
const pair<pair<long double, long double>, long double> CFAIL =
    pair<pair<long double, long double>, long double>(FAIL, 0);
pair<pair<long double, long double>, long double> incircle(
    const pair<long double, long double> &a,
    const pair<long double, long double> &b,
    const pair<long double, long double> &c) {
  if (is_zero(cp(b - a, c - a))) return CFAIL;
  pair<pair<long double, long double>, pair<long double, long double> > p =
      segment_bisector(
          pair<pair<long double, long double>, pair<long double, long double> >(
              a, b));
  pair<pair<long double, long double>, pair<long double, long double> > q =
      segment_bisector(
          pair<pair<long double, long double>, pair<long double, long double> >(
              a, c));
  pair<long double, long double> t = intersection_line_line(p, q);
  return pair<pair<long double, long double>, long double>(t, pieprzu(t, a));
}
pair<long double, long double> intersection_segment_segment(
    const pair<pair<long double, long double>, pair<long double, long double> >
        &u,
    const pair<pair<long double, long double>, pair<long double, long double> >
        &v) {
  pair<long double, long double> r = intersection_line_line(u, v);
  if (on_segment(r, u) and on_segment(r, v))
    return r;
  else
    return FAIL;
}
int main() {
  int N, X, Y;
  scanf("%d %d %d", &N, &X, &Y);
  vector<pair<long double, long double> > P;
  int max_a = 0;
  int max_b = 0;
  for (int(i) = (1); (i) <= (N); (i)++) {
    int a, b;
    scanf("%d %d", &a, &b);
    max_a = max(max_a, a);
    max_b = max(max_b, b);
    P.emplace_back(a, b);
  }
  P.emplace_back(max_a, 0);
  P.emplace_back(0, max_b);
  P = convex_hull(P);
  pair<long double, long double> n(X, Y);
  pair<long double, long double> z;
  for (int(i) = (1); (i) <= (1500000); (i)++) {
    if (X * i > 1490490 or Y * i > 1490010) {
      z = pair<long double, long double>(X * i, Y * i);
      break;
    }
  }
  pair<pair<long double, long double>, pair<long double, long double> > s(
      pair<long double, long double>(0, 0), z);
  long double result = INF;
  for (int(i) = (0); (i) < (int(P.size())); (i)++) {
    pair<long double, long double> a = P[i];
    pair<long double, long double> b = P[(i + 1) % int(P.size())];
    pair<pair<long double, long double>, pair<long double, long double> > t(a,
                                                                            b);
    pair<long double, long double> d = intersection_segment_segment(s, t);
    if (d == FAIL) continue;
    result = min(result, pieprzu(pair<long double, long double>(X, Y),
                                 pair<long double, long double>(0, 0)) /
                             pieprzu(d, pair<long double, long double>(0, 0)));
  }
  cout << fixed << result << endl;
  return 0;
}