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.
//package codeforces.cfr335div1;

import java.io.*;
import java.util.*;
import java.util.function.Function;

/**
 * Created by raggzy on 4/2/2016.
 */
public class C {

    private static class Reader implements Closeable {
        private BufferedReader br;
        private StringTokenizer st;

        public Reader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        public void readLine() {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

        public String next() {
            if (st == null || !st.hasMoreTokens()) {
                readLine();
            }
            return st.nextToken();
        }

        public <T> T next(Function<String, T> parser) {
            return parser.apply(next());
        }

        public int nextInt() {
            return next(Integer::valueOf);
        }

        public long nextLong() {
            return next(Long::valueOf);
        }

        public double nexDouble() {
            return next(Double::valueOf);
        }

        @Override
        public void close() throws IOException {
            br.close();
        }
    }

    private static class Pair<A, B> {
        protected final A a;
        protected final B b;

        public Pair(A a, B b) {
            this.a = a;
            this.b = b;
        }
    }

    private static class Point extends Pair<Long, Long> implements Comparable<Point> {
        public Point(Long a, Long b) {
            super(a, b);
        }

        @Override
        public int compareTo(Point o) {
            int c1 = compareAngle(o);
            int c2 = Long.compare(a, o.a);
            return c1 != 0 ? c1 : c2;
        }

        private int compareAngle(Point o) {
            return Long.compare(b * o.a, a * o.b);
        }

        private static long vecMul(Point p1, Point p2, Point p3) {
            long x1 = p1.a - p2.a;
            long y1 = p1.b - p2.b;
            long x2 = p3.a - p2.a;
            long y2 = p3.b - p2.b;
            return x1 * y2 - x2 * y1;
        }

        private static Pair<Double, Double> lineIntersection(Point l1p1, Point l1p2, Point l2p1, Point l2p2) {
            long A1 = l1p1.b - l1p2.b;
            long B1 = l1p2.a - l1p1.a;
            long C1 = l1p1.a * l1p2.b - l1p2.a * l1p1.b;
            long A2 = l2p1.b - l2p2.b;
            long B2 = l2p2.a - l2p1.a;
            long C2 = l2p1.a * l2p2.b - l2p2.a * l2p1.b;
            long det = A1 * B2 - A2 * B1;
            double a = 1.0 * (C2 * B1 - C1 * B2) / det;
            double b = 1.0 * (C1 * A2 - C2 * A1) / det;
            return new Pair<>(a, b);
        }
    }

    public static void main(String[] args) {
        Reader in = new Reader(System.in);
        int n = in.nextInt();
        long p = in.nextInt();
        long q = in.nextInt();
        List<Point> points = new ArrayList<>(n + 2);
        long maxA = Long.MIN_VALUE;
        long maxB = Long.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            Point e = new Point(in.nextLong(), in.nextLong());
            maxA = Math.max(maxA, e.a);
            maxB = Math.max(maxB, e.b);
            points.add(e);
        }
        points.add(new Point(maxA, 0L));
        points.add(new Point(0L, maxB));
        // sort by angle
        Collections.sort(points);
        // build convex hull
        LinkedList<Point> hull = new LinkedList<>();
        for (Point curr : points) {
            if (hull.size() < 2) {
                hull.add(curr);
            } else {
                while (hull.size() >= 2) {
                    Point p1 = hull.get(hull.size() - 2);
                    Point p2 = hull.getLast();
                    long mul = Point.vecMul(p1, p2, curr);
                    if (mul < 0) break;
                    hull.removeLast();
                }
                hull.add(curr);
            }
        }
        Point zero = new Point(0L, 0L);
        Point target = new Point(p, q);
        Point prev = null;
        for (Point curr : hull) {
            if (curr.compareAngle(target) >= 0) {
                // this is exactly the point, just take prev segment and intersect with line (p,q)
                Pair<Double, Double> pt = Point.lineIntersection(zero, target, curr, prev);
                System.out.println(Math.sqrt(p * p + q * q) / Math.sqrt(pt.a * pt.a + pt.b * pt.b));
                return;
            }
            prev = curr;
        }
    }
}
