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>
using namespace std;
#define RI(x) scanf("%d",&(x))
#define RII(x,y) scanf("%d %d",&(x),&(y))
#define RI64(x) scanf("%I64d",&(x))
#define RII64(x,y) scanf("%I64d%I64d",&(x),&(y))
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(),(x).end()
#define pritnf printf
#define N 200514
using namespace std;
typedef long long int lnt;
typedef double dou;
const dou eps=1e-12;
int sgn(dou k){return (k>-eps)-(k<eps);}
typedef struct{dou x,y;int id;}dot;

dot operator-(dot a,dot b){return(dot){a.x-b.x,a.y-b.y};}
dot operator+(dot a,dot b){return(dot){a.x+b.x,a.y+b.y};}
dot operator*(dou k,dot a){return(dot){a.x*k,a.y*k};}

dou operator*(dot a,dot b){return a.x*b.y-a.y*b.x;}
dou atan2(dot a){return atan2(a.y,a.x);}
int cmp(dot a,dot b){
	return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
int cmp2(dot a,dot b){return a.id<b.id;}
dot d[N],tb[N];int ct,dt;
int n,bad[N];
map<pair<int,int>,int>mp;
void sol(){
	mp.clear();
	dt=0;
	for(int i=0;i<n;i++){
		bad[i]=0;
		int x,y;
		RII(x,y);
		pair<int,int>pp=pair<int,int>(x,y);
		int t=mp[pp];
		if(t){
			bad[t]=1;
			continue;
		}
		d[dt++]=(dot){10000/x,10000/y,i+1};
		mp[pp]=i+1;
	}
	sort(d,d+dt,cmp);
	int st=0;
	ct=0;
	for(int i=dt-1;i>=0;i--){
		for(;ct>=2&&sgn((tb[ct-1]-tb[ct-2])*(d[i]-tb[ct-1]))>=0;ct--);
		tb[ct++]=d[i];
	}
	for(;ct-st>=2&&sgn(atan2(tb[st]-tb[st+1]))>=0;st++);
	for(;ct-st>=2&&sgn(atan2(tb[ct-1]-tb[ct-2])-acos(-1)/2.0)<=0;ct++);
	/////////////////////
	sort(tb+st,tb+ct,cmp2);
	for(int i=st;i<ct;i++)if(!bad[tb[i].id])printf("%d ",tb[i].id);
	puts("");
}
int main(){
	RI(n);
	sol();
	return 0;
}

