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.
import java.util.*;
import java.io.*;
import java.text.*;
public class Main{
    //SOLUTION BEGIN
    //Into the Hardware Mode
    void pre() throws Exception{}
    void solve(int TC) throws Exception {
        int n = ni();long a = nl(), b = nl();
        Point[] p = new Point[n];
        for(int i = 0; i< n; i++)p[i] = new Point(ni(), ni());
        p = convexHull(p);
        n = p.length;
        if(n == 1){
            pn(df.format(Math.max(a*1.0/p[0].x, b*1.0/p[0].y)));
            return;
        }
        double ans = Double.MAX_VALUE;
        for(int i = 0; i< n; i++){

            ans = Math.min(ans, Math.max(a*1.0/p[i].x, b*1.0/p[i].y));
            Point A = p[i], B = p[(i+1)%n];
            if(A.x*B.y == B.x*A.y)continue;
            double g = ((A.y*a*1.0)/A.x-b)/(B.x*A.y*1.0/A.x-B.y);
            double f = (b-B.y*g)/A.y;
            if(Math.min(f, g) > eps)ans = Math.min(ans, f+g);
        }
        pn(df.format(ans));
    }
    Point[] convexHull(Point[] p){
        int ind = 0;
        for(int i = 1; i< p.length; i++)
            if(p[i].y < p[ind].y || (p[i].y == p[ind].y && p[i].x < p[ind].x))ind = i;
        {
            Point tmp = p[ind];
            p[ind] = p[0];
            p[0] = tmp;
        }
        Arrays.sort(p, 1, p.length, (Point p1, Point p2) -> {
            int o = ccw(p[0], p1, p2);
            if(o == 0)return Long.compare(p[0].distSq(p1), p[0].distSq(p2));
            return o;
        });
        Point[] ans = new Point[p.length];
        int sz = 0;
        ans[sz++] = p[0];
        for(int i = 1; i< p.length; i++){
            while(sz > 1 && ccw(ans[sz-2], ans[sz-1], p[i]) >= 0)sz--;
            ans[sz++] = p[i];
        }
        return Arrays.copyOf(ans, sz);
    }
    int ccw(Point p, Point q, Point r){
        long val = (q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y);
        if(val == 0)return 0;
        return val > 0?1:-1;
    }
    class Point{
        long x, y;
        public Point(long X, long Y){
            x = X;y = Y;
        }
        long distSq(Point p){
            return (x-p.x)*(x-p.x)+(y-p.y)*(y-p.y);
        }
    }
    void dfs2(int[][] g, int[] ans, int[] val, int u, int p){
        ans[u] = val[u];
        for(int v:g[u]){
            if(v == p)continue;
            val[u] -= Math.max(val[v], 0);
            val[v] += Math.max(val[u], 0);
            dfs2(g, ans, val, v, u);
            val[v] -= Math.max(val[u], 0);
            val[u] += Math.max(val[v], 0);
        }
    }
    void dfs(int[][] g, int[] val, int[] a, int u, int p){
        val[u] = a[u];
        for(int v:g[u]){
            if(v == p)continue;
            dfs(g, val, a, v, u);
            val[u] += Math.max(val[v], 0);
        }
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    void exit(boolean b){if(!b)System.exit(0);}
    //    long IINF = (long)1e15;
    final int INF = (int)1e9+2;//, MX = (int)2e6+5;
    DecimalFormat df = new DecimalFormat("0.00000000000");
    double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-8;
    static boolean multipleTC = false, memory = false, fileIO = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        long ct = System.currentTimeMillis();
        if (fileIO) {
            in = new FastReader("");
            out = new PrintWriter("");
        } else {
            in = new FastReader();
            out = new PrintWriter(System.out);
        }
        //Solution Credits: Taranpreet Singh
        int T = (multipleTC) ? ni() : 1;
        pre();
        for (int t = 1; t <= T; t++) solve(t);
        out.flush();
        out.close();
        System.err.println(System.currentTimeMillis() - ct);
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    int[][] make(int n, int[] from, int[] to, int e, boolean f){
        int[][] g = new int[n][];int[]cnt = new int[n];
        for(int i = 0; i< e; i++){
            cnt[from[i]]++;if(f)cnt[to[i]]++;
        }
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
        for(int i = 0; i< e; i++){
            g[from[i]][--cnt[from[i]]] = to[i];
            if(f)g[to[i]][--cnt[to[i]]] = from[i];
        }
        return g;
    }
    int[][][] makeS(int n, int[] from, int[] to, boolean f){
        int[][][] g = new int[n][][];int[]cnt = new int[n];
        for(int i:from)cnt[i]++;if(f)for(int i:to)cnt[i]++;
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]][];
        for(int i = 0; i< from.length; i++){
            g[from[i]][--cnt[from[i]]] = new int[]{to[i], i};
            if(f)g[to[i]][--cnt[to[i]]] = new int[]{from[i], i};
        }
        return g;
    }
    int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
    int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception{
            String str;
            try{
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }
            return str;
        }
    }
}