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 JAVA):

Solution:
import java.io.*;
import java.util.*;
public class c {
public static void main(String[] args) throws IOException {
    input.init(System.in);
    PrintWriter out = new PrintWriter(System.out);
    int n = input.nextInt();
    long p = input.nextInt(), q = input.nextInt();
    Point[] data = new Point[n];
    for(int i = 0; i<n; i++) data[i] = new Point(input.nextInt(), input.nextInt());
    Point[] hull = findHull(data);
    double min = 1e9;
    n = hull.length;
    //System.out.println(n);
    for(int i = 0; i<n; i++)
    {
        Segment cur = new Segment(hull[i], hull[(i+1)%n]);
        Segment go = new Segment(new Point(0, 0), new Point(p, q));
        //System.out.println(cur.start+" "+cur.end);
        Point inter = cur.start.compareTo(cur.end) == 0 ? cur.start : segmentIntersection(cur, go);
        if(inter != null)
        {
            //System.out.println(inter.x+" "+inter.y);
            double time = Math.max(p / inter.x, q/inter.y);
            min = Math.min(min,  time);
        }
    }
    for(int i = 0; i<n; i++)
    {
        double max = Math.max(p*1.0 / hull[i].x, q*1.0/hull[i].y);
        min = Math.min(min,  max);
    }
    out.println(min);
    out.close();
}
public static class Segment
{
    Point start, end;
    public Segment(Point a, Point b)
    {
        start = a; end = b;
    }
}
static double EPS = 1e-6;
public static Point segmentIntersection(Segment s1, Segment s2)
{
    Point p = lineIntersection(s1, s2);
    if(p==null)
    {
        //segments are parallel
        return null;
    }
    if(!(p.x >= Math.min(s1.start.x,s1.end.x)-EPS && p.x <= Math.max(s1.end.x,s1.start.x)+EPS))
        return null;
    //if(!(p.x >= Math.min(s2.start.x,s2.end.x)-1E-9 && p.x <= Math.max(s2.end.x,s2.start.x)+1E-9))
        //return null;
    if(!(p.y >= Math.min(s1.start.y,s1.end.y)-EPS && p.y <= Math.max(s1.end.y,s1.start.y)+EPS))
        return null;
    //if(!(p.y >= Math.min(s2.start.y,s2.end.y)-1E-9 && p.y <= Math.max(s2.end.y,s2.start.y)+1E-9))
    //  return null;
    return p;
}
public static Point lineIntersection(Segment s1, Segment s2)
{
    double a1 = s1.end.y - s1.start.y;
    double b1 = s1.start.x - s1.end.x;
    double c1 = a1*s1.start.x + b1*s1.start.y;
    double a2 = s2.end.y - s2.start.y;
    double b2 = s2.start.x - s2.end.x;
    double c2 = a2*s2.start.x + b2*s2.start.y;
    double det = a1*b2 - a2*b1;
    if(det == 0)
    {
        //lines are parallel
        return null;
    }
    double x = (b2*c1 - b1*c2)/det;
    double y = (a1*c2 - a2*c1)/det;
    if(x == -0) x=0;
    if(y == -0) y=0;
    return new Point(x, y);
}
static class Point implements Comparable<Point>
{
    double x, y;
    public Point(double xx, double yy)
    {
        x = xx; y = yy;
    }
    @Override
    public int compareTo(Point o) {
        if(Math.abs(x - o.x)  > 1e-9) return (int)Math.signum(x - o.x);
        return (int)Math.signum(y - o.y);
    }
    Point minus(Point p)
    {
        return new Point(x - p.x, y - p.y);
    }
    double cross(Point p)
    {
        return x * p.y - y*p.x;
    }
    public String toString()
    {
        return x+" "+y;
    }
}
//Monotone Chain Convex Hull Algorithm
//O(nlogn), where n is the number of points
//returns all vertices and intermediate points of the hull in counter-clockwise order
//Point class needs minus(Point that), comparator of increasing x then y, and cross(Point that)
public static Point[] findHull( Point[] points)
{
int n = points.length;
Arrays.sort( points);
Point[] ans = new Point[2 * n];             // In between we may have a 2n points
int k = 0, start = 0;                   // start is the first insertion point 
for(int i = 0; i < n; i ++)                     // Finding lower layer of hull
{
    Point p = points[i];
    while( k - start >= 2 && p.minus( ans[k-1] ).cross(p.minus( ans[k-2] ) ) >= 0 )
        k--;
    ans[k++] = p; 
}                       // drop off last point from lower layer
start = k ;                     
for(int i = n-1 ; i >= 0 ; i --)                // Finding top layer from hull
{
    Point p = points[i];
    while( k - start >= 2 && p.minus( ans[k-1] ).cross(p.minus( ans[k-2] ) ) >= 0 )
        k--;
    ans[k++] = p; 
}
return Arrays.copyOf(ans, k);                   // convex hull is of size k
}
public static class input {
    static BufferedReader reader;
    static StringTokenizer tokenizer;

    static void init(InputStream input) {
    reader = new BufferedReader(new InputStreamReader(input));
    tokenizer = new StringTokenizer("");
    }

    static String next() throws IOException {
    while (!tokenizer.hasMoreTokens())
        tokenizer = new StringTokenizer(reader.readLine());
    return tokenizer.nextToken();
    }

    static int nextInt() throws IOException {
    return Integer.parseInt(next());
    }

    static double nextDouble() throws IOException {
    return Double.parseDouble(next());
    }

    static long nextLong() throws IOException {
    return Long.parseLong(next());
    }
}
}