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;
typedef pair<double ,double > dd;
double intersect(dd x,dd y)
{
    if(x.first==y.first) return INFINITY;
    return (double)(y.second-x.second)/(x.first-y.first);
}
struct convex_hull
{
    dd b[300000];
    int siz;
    convex_hull()
    {
        siz=0;
    }
    double cal(double x,int y)
    {
        return b[y].first*x+b[y].second;
    }
    void insert(double xx,double yy)
    {
        //if(siz>0)printf("%lf %lf")
        while(siz>0 && b[siz].first==xx && b[siz].second>yy) siz--;
        while(siz>1 && intersect(b[siz-1],dd(xx,yy))<intersect(b[siz-1],b[siz])) siz--;
        b[++siz]=dd(xx,yy);
        //printf("%lf %lf %d\n",xx,yy,siz);
    }
}convex;
vector<dd> a;
vector<int> ans;
map<dd,int> ma;
int next[300000];
int main()
{
    double Y=200000000;
    int n,i;
    //freopen("test.inp","r",stdin);
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        double vx,vy;
        scanf("%lf %lf",&vx,&vy);
       // printf("%d %lf %lf\n",i,1/vx,Y/vy);
        a.push_back(dd(1000000/vx,Y/vy));
        if(ma.find(dd(1000000/vx,Y/vy))!=ma.end())
        {
            next[i]=ma[dd(1000000/vx,Y/vy)];
        }
        ma[dd(1000000/vx,Y/vy)]=i;
    }
    sort(a.begin(),a.end());
    reverse(a.begin(),a.end());
    for(i=0;i<a.size();i++)
    {
        convex.insert(a[i].first,a[i].second);
    }
    for(i=1;i<=convex.siz;i++)
    {
        if(i!=convex.siz && convex.cal(0.0001,i)>convex.cal(0.0001,i+1)) continue;
        //printf("%d %lf %lf %d\n",i,convex.b[i].first,convex.b[i].second,ma[convex.b[i]]);
        int cur=ma[convex.b[i]];
        ans.push_back(cur);
        while((cur=next[cur]) && cur!=0)
        ans.push_back(cur);
    }
    sort(ans.begin(),ans.end());
    for(i=0;i<ans.size();i++) printf("%d ",ans[i]);
}
