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<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<list>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#include<string>
#define REP(it,end) for (int it = 1; it <= (end); it++)
#define FOR(it,begin,end) for (int it = (begin); it <= (end); it++)
#define ROF(it,begin,end) for (int it = (begin); it >= (end); it--)
using namespace std;
const int maxn=2e5+100;
typedef long long ll;
struct Point
{
    ll x,y;
    bool operator <(const Point &p)const
    {
        if(x!=p.x)return x<p.x;
        else return y<p.y;
    }
}pt[maxn];
int stk[maxn],stnum;
set<pair<ll,ll> >has;
bool check(Point a,Point b,Point c)
{
    return c.x*a.y*(b.x-a.x)*(b.y-c.y)>=a.x*c.y*(b.x-c.x)*(b.y-a.y);
}
void convex(int n)
{
    int i,j;
    sort(pt,pt+n);
    stnum=0;
    for(i=0;i<n;i++)
    {
        if(stnum>1&&check(pt[stk[stnum-1]],pt[stk[stnum-2]],pt[i]))stnum--;
        stk[stnum++]=i;
    }
    for(i=0;i<stnum;i++)
        has.insert(make_pair(pt[i].x,pt[i].y));
}
ll a[maxn],b[maxn];
pair<ll,ll> node[maxn];
int main()
{
    int n,i,j;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%I64d %I64d",&a[i],&b[i]);
        node[i]=make_pair(a[i],b[i]);
    }
    sort(node,node+n);
    for(i=0,j=0;i<n;i++)
    {
        while(j>0&&node[j-1].second<=node[i].second)j--;
        node[j++]=node[i];
    }
    for(i=0;i<j;i++)
    {
        pt[i].x=node[i].first;
        pt[i].y=node[i].second;
    }
    convex(j);
    for(i=0;i<n;i++)
    {
        //cout <<a[i]<<" "<<b[i]<<endl;
        if(has.find(make_pair(a[i],b[i]))!=has.end())
            printf("%d ",i+1);
    }
    printf("\n");
    return 0;
}
