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

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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 For(i,a,b) for(long long i=a;i<b;i++)
#define pb push_back
#define mod 1000000007
#define reset(s,val) memset(s,val,sizeof(s))
#define eps 0.0000001
#define pi acos(-1)
#define sqr(x) (x)*(x)
#define maxn 32000

struct point
{
    double x,y,agl;
    int st;
};

int n,s,r,x0,y0,ss,rr,bef[200005];
point lis[200005],p0,p1,pp;
stack<int> S;
vector<int> ans;
vector<point> liss;

double dis(point p1,point p2)
{
    return sqr(p1.x-p2.x)+sqr(p1.y-p2.y);
}

bool com(point p1,point p2)
{
    return (p1.agl<p2.agl||(p1.agl==p2.agl&&dis(p1,p0)<dis(p2,p0)));
}

bool oppo(point p1,point p2,point p3,point p4)
{
    return (((p1.y-p2.y)*(p3.x-p1.x)+(p2.x-p1.x)*(p3.y-p1.y))*((p1.y-p2.y)*(p4.x-p1.x)+(p2.x-p1.x)*(p4.y-p1.y))<0);
}

int main( ){
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    cin>>n;
    p0.x=p0.y=p1.x=p1.y=100000.0;
    For(i,0,n)
    {
        cin>>s>>r;
        lis[i].x=1.0/s;
        lis[i].y=1.0/r;
        lis[i].st=i+1;
        lis[i].agl=atan2(lis[i].x,lis[i].y);
        if(lis[i].x<p0.x||lis[i].x==p0.x&&lis[i].y<p0.y) p0=lis[i];
        if(lis[i].y<p1.y||lis[i].y==p1.y&&lis[i].x<p1.x) p1=lis[i];
    }
    pp.x=pp.y=0;
    For(i,1,n) if(!oppo(p0,p1,lis[i],pp)) liss.pb(lis[i]);
    sort(liss.begin(),liss.end(),com);
    S.push(0);S.push(1);
    For(i,2,n)
    {
        while(S.top()!=1)
        {
            int cur=S.top();
            if(oppo(lis[i],lis[bef[cur]],lis[cur],lis[0])) S.pop();
            else break;
        }
        bef[i]=S.top();
        S.push(i);
    }

    ans.clear();
    //cout<<S.size();
    //cout<<p0.st<<endl;
    while(!S.empty())
    {
        if(!oppo(p0,p1,lis[S.top()],pp)) ans.pb(lis[S.top()].st);
        S.pop();
        //cout<<ans.size()<<endl;
    }
    sort(ans.begin(),ans.end());
    for(auto i:ans) cout<<i<<' ';
}