Tavas is a cheerleader in the new sports competition named "Pashmaks".

<image>

This competition consists of two part: swimming and then running. People will immediately start running R meters after they finished swimming exactly S meters. A winner is a such person that nobody else finishes running before him/her (there may be more than one winner).

Before the match starts, Tavas knows that there are n competitors registered for the match. Also, he knows that i-th person's swimming speed is si meters per second and his/her running speed is ri meters per second. Unfortunately, he doesn't know the values of R and S, but he knows that they are real numbers greater than 0.

As a cheerleader, Tavas wants to know who to cheer up. So, he wants to know all people that might win. We consider a competitor might win if and only if there are some values of R and S such that with these values, (s)he will be a winner.

Tavas isn't really familiar with programming, so he asked you to help him.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 2 × 105).

The next n lines contain the details of competitors. i-th line contains two integers si and ri (1 ≤ si, ri ≤ 104).

Output

In the first and the only line of output, print a sequence of numbers of possible winners in increasing order.

Examples

Input

3
1 3
2 2
3 1


Output

1 2 3 


Input

3
1 2
1 1
2 1


Output

1 3 
5
10 50
10 50
10 50
10 50
10 50
 2
10000 10000
10000 10000
 3
10000 10000
10000 10000
10000 10000
 3
1000 3000
1500 1500
3000 1000
 43
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 1
1 1
 2
1 1
1 1
 44
78 52
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 3
1000 3000
1499 1499
3000 1000
 5
50 50
49 50
50 49
49 49
50 1
 18
82 38
33 69
33 69
33 69
33 69
33 69
82 38
33 69
33 69
33 69
82 38
33 69
33 69
82 38
82 38
33 69
33 69
33 69
 1
10000 9999
 3
1000 3000
1501 1501
3000 1000
 2
1 2
2 1

1 2 3 4 5
 1 2
 1 2 3
 1 2 3
 1 3 16 24 29 31 32 37 41 42
 1
 1 2
 1
 1 3
 1
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
 1
 1 2 3
 1 2 

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define fore(i,a,b) for(int i=a,ThxDem=b;i<ThxDem;++i)
using namespace std;
typedef long long ll;

const double EPS=1e-9;

struct pt {  // for 2D ignore z coordinate
	double x,y;int id;
	pt(double x, double y, int id) : x(x),y(y),id(id){}
	pt(double x, double y): x(x),y(y){}
	pt(){}
	double norm(){return sqrt(*this**this);}
	pt operator+(pt p){return pt(x+p.x,y+p.y);}
	pt operator-(pt p){return pt(x-p.x,y-p.y);}
	pt operator*(double t){return pt(t*x,t*y);}
	double operator*(pt p) {return x*p.x+y*p.y;}
//	pt operator^(pt p){ // only for 3D
//		return pt(y*p.z-z*p.y,z*p.x-x*p.z,x*p.y-y*p.x);}
//	double operator%(pt p){return (*this^p).norm();}
	double operator%(pt p){return x*p.y-y*p.x;} // for 2D
	bool operator<(pt p){ // for convex hull
		return x<p.x-EPS||(abs(x-p.x)<EPS&&y<p.y-EPS);}
	bool left(pt p, pt q){ // is it to the left of directed line pq?
		return (p-*this)%(q-*this)>EPS;}
};
// CCW order
// Includes collinear points (change sign of EPS in left to exclude)
vector<pt> chull(vector<pt> ps){
	vector<pt> r;
	sort(ps.begin(),ps.end()); // first x, then y
	fore(i,0,ps.size()){ // lower hull
		while(r.size()>=2&&r.back().left(r[r.size()-2],ps[i]))r.pop_back();
		r.pb(ps[i]);
	}
	r.pop_back();
	int k=r.size();
	for(int i=ps.size()-1;i>=0;--i){ // upper hull
		while(r.size()>=k+2&&r.back().left(r[r.size()-2],ps[i]))r.pop_back();
		r.pb(ps[i]);
	}
	r.pop_back();
	return r;
}

bool r[200005];
int n;
int a[200005],b[200005],ia,ib;
vector<pt> p;
pair<int,int> ma,mb;

int main(){
	scanf("%d",&n);
	fore(i,0,n)scanf("%d%d",a+i,b+i),ma=max(ma,mp(a[i],b[i])),mb=max(mb,mp(b[i],a[i]));
	fore(i,0,n){
		if(mp(a[i],b[i])==ma)r[i]=true,ia=i;
		if(mp(b[i],a[i])==mb)r[i]=true,ib=i;
		p.pb(pt(1./a[i],1./b[i],i));
	}
	vector<pt> c=chull(p);
	int k=0;
	for(;c[k].id!=ia;++k);
	for(;c[k].id!=ib;++k,k%=c.size())r[c[k].id]=true;
	bool s=false;
	fore(i,0,n)if(r[i]){
		if(s)putchar(' ');s=true;
		printf("%d",i+1);
	}
	puts("");
	return 0;
}