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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class C335Div1CFreDre {
	static final double EPS = 1e-9;
	
	static int N;
	static double A,B;
	static DoublePoint[] P;
	static Line[] hullLines;
	
	public static void main(String args[]){
//		prln(EPS+ "");
		BufferedReader br  = new BufferedReader(new InputStreamReader(System.in));
		
		try{
			String[] nab = br.readLine().split(" ");
			N  = Integer.parseInt(nab[0]);
			A = Double.parseDouble(nab[1]);
			B = Double.parseDouble(nab[2]);
			P = new DoublePoint[N+2];
			double maxX = 0, maxY = 0;
			double x, y;
			String[] lns;
			for(int i = 0; i < N; i++){
				lns = br.readLine().split(" ");
				x =  Double.parseDouble(lns[0]);
				y = Double.parseDouble(lns[1]);
				maxX = Double.max(maxX, x);
				maxY = Double.max(maxY, y);
				P[i+1] = new DoublePoint(x, y);
			}
			P[0] = new DoublePoint(0, maxY + EPS/100);
			P[N+1] = new DoublePoint(maxX+EPS/100, 0);
			Arrays.sort(P);
//			prln(Arrays.toString(P));
			DoublePoint[] cvHull = new DoublePoint[N+2];
			int nHull = 0;
			cvHull[0] = P[0];
			int k = 2;
			for(; k < N+1; k++) if(Math.abs(P[k].x - P[1].x) > EPS) break;
			k--;
			cvHull[1]= P[k];
//			k++;
//			for(; k < N+1; k++){
//				if(Math.abs(P[k].x - P[1].x) > EPS && P[k].y > P[1].y+ EPS) break;
//			}
//			System.out.println("k = " + k);
//			cvHull[2] = P[k];
			nHull = 2;
			hullLines = new Line[N+1];
			hullLines[0] = new Line(cvHull[nHull-1].x, cvHull[nHull-1].y, cvHull[nHull-2].x, cvHull[nHull-2].y);			
//			hullLines[1] = new Line(cvHull[nHull-1].x, cvHull[nHull-1].y, cvHull[nHull-2].x, cvHull[nHull-2].y);			
			int nlines = 1;			
			int pos;			
			int i = k+1;
			while(i < N+1){
				int j = i+1;
				for(; j < N+1; j++)if(Math.abs(P[j].x-P[i].x) > EPS)break;
				j--;
//				prln("j: " + j);
				if(P[j].y > cvHull[nHull-1].y + EPS) {
					pos = hullLines[nlines -1].relativePos(P[j].x, P[j].y);
//					prln("check " +j + " pos: "+ pos);
					if(pos > 0){
						int lineNo = searchLine(P[j], nlines-1,0);
						cvHull[lineNo+2] = P[j];
						nHull = lineNo+3;
						hullLines[lineNo+1] = new Line(P[j].x, P[j].y, cvHull[lineNo+1].x, cvHull[lineNo+1].y);
						nlines = lineNo+2;
//						prln("add " + P[i].toString());
					}else {
							cvHull[nHull] = P[j];
							nHull++;
							hullLines[nlines] = new Line(cvHull[nHull-1].x, cvHull[nHull-1].y, cvHull[nHull-2].x, cvHull[nHull-2].y);
							nlines++;
//							prln("add " + P[i].toString());
						}
				}
				i = j +1;
			}
//			if(k != N+1){
			cvHull[nHull] = P[N+1];
			nHull++;
//			}
			Line optLine = new Line(0, 0, A, B);
			int tmp, prev;
			prev = optLine.relativePos(cvHull[0].x, cvHull[0].y);
			i = 0;
//			prln("nhull " + nHull);
			for(i = 1; i < nHull; i++){
				tmp = optLine.relativePos(cvHull[i].x, cvHull[i].y);
				if(tmp != prev) break;
				else prev = tmp;
			}
//			System.out.println(i);
			DoublePoint intersect = optLine.intersect(new Line(cvHull[i-1].x, cvHull[i-1].y, cvHull[i].x, cvHull[i].y));
//			prln(cvHull[0].toString());
//			prln(cvHull[i-1].toString());
//			prln(cvHull[i].toString());
//			prln(intersect.toString());
			double time = A/intersect.x;
			prln(time + "");
//			System.out.println(nHull);
//			prln(Arrays.toString(cvHull));
			
//			prln(Arrays.toString(P));
			
		}catch(IOException ioe){
			ioe.printStackTrace();
		}
	}
	
//	static int searchLine(DoublePoint P, int h, int l){
////		prln("search");
//		if(h < l) return -1;
//		else if(h == l) return h;
//		else{
//			int m =  (h+l)/2;
//			int pos  = hullLines[m].relativePos(P.x, P.y);
//			if(pos < 0) {
//				int posp = hullLines[m+1].relativePos(P.x, P.y);
//				if(posp < 0) return searchLine(P, m+1, h);
//				else return m;
//			}
//			else return searchLine(P, l, m-1);
//		}
//	}
	
	// return the largest k  where P is below hullLinek[k] 
	static int searchLine(DoublePoint P, int h, int l){
//		prln("start search = " + h + " point:" + P.toString());
		int pos = 1;
		while(pos > 0){
//			prln(hullLines[h].toString());
			pos = hullLines[h].relativePos(P.x, P.y);
			h--;
		}
		return h+1;
		
//		prln("search");
//		if(h < l) return -1;
//		else if(h == l) return h;
//		else{
//			int m =  (h+l)/2;
//			int pos  = hullLines[m].relativePos(P.x, P.y);
//			if(pos < 0) {
//				int posp = hullLines[m+1].relativePos(P.x, P.y);
//				if(posp < 0) return searchLine(P, m+1, h);
//				else return m;
//			}
//			else return searchLine(P, l, m-1);
//		}
	}
	static void prln(String s){
		System.out.println(s);
	}

	
	static class Line{
		private double xa,xb;
		private double ya,yb;
		
		public Line(double xa, double ya, double xb, double yb) {
			this.xa = xa; this.xb  = xb;
			this.ya = ya; this.yb = yb;
		}
		
		public int relativePos(double x, double y){
			if(Math.abs(xa -xb) < EPS){
				return (Math.abs(x-xa) < EPS ? 0: (x < xa ? -1: 1));
			} else if(Math.abs(ya -yb) < EPS){
				return (Math.abs(y-ya) < EPS ? 0: (y < ya ? -1: 1));
			} else {
				double rs = (y-ya)/(ya-yb) - (x-xa)/(xa-xb);
				return (Math.abs(rs) < EPS? 0: (rs < 0? -1 : 1));
			}
		}
		
		public boolean isParallelX(){
			if(Math.abs(ya - yb) < EPS) return true;
			else return false;
		}
		
		public boolean isPrallelY(){
			if(Math.abs(xa - xb) < EPS) return true;
			else return false;
		}
		
		public DoublePoint intersect(Line l){
			if(l.isParallelX()){
				return new DoublePoint(xa + (xa - xb)*(l.ya-ya)/(ya-yb), l.ya);
			}else if(l.isPrallelY()){
				return new DoublePoint(l.xa, ya + (ya-yb)*(l.xa -xa)/(xa-xb));
			}else {
				double a1 = (ya-yb)/(xa-xb);
				double b1 = ya - (ya-yb)/(xa-xb)*xa;
				double a2 = (l.ya-l.yb)/(l.xa-l.xb);
				double b2  = l.ya - (l.ya-l.yb)/(l.xa-l.xb)*l.xa;
				return new DoublePoint((b2-b1)/(a1-a2), a1*(b2-b1)/(a1-a2) + b1);
			}
		}
		
		public String toString(){
			return "[(" + xa + "," + ya + ")," + "(" + xb + "," + yb + ")]";
		}
		
	}
	
	static class DoublePoint implements Comparable<DoublePoint>{
		private double x; 
		private double y;
		
		public DoublePoint(double x, double y) {
			this.x  = x;
			this.y  = y;
		}

		@Override
		public int compareTo(DoublePoint o) {
			if(Math.abs(o.x-x) < EPS){
				return Double.compare(y, o.y);
			}else return Double.compare(o.x, x);
		}
		
		public String toString(){
			return "(" + x + "," + y + ")";
		}
		
		
	}
}

//10 630 764
//679 16
//34 691
//778 366
//982 30
//177 9
//739 279
//992 89
//488 135
//7 237
//318 318
//
//output: 1.472265278375486


//10 123457 123459
//5 2
//11 4
//1 8
//11 1
//7 1
//10 2
//8 1
//11 3
//3 4
//11 8

//output: 15432.375000000000000