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<algorithm>
#include<bitset>
#include<cctype>
#include<climits>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<string>
#include<vector>

using namespace std;

const int maxn=200001;

typedef long long LL;
typedef long double LD;
typedef int array[maxn];

struct Line
{
	double k,b;
	int id;
	Line(){}
	Line(double k,double b,int id):k(k),b(b),id(id){}
	friend bool operator < (Line x,Line y)
	{
		return x.b==y.b?x.k<y.k:x.b<y.b;
	}
}l[maxn],st[maxn];
int n,m,top;
array ans;

bool check(Line l1,Line l2,Line l3)
{
	double x=(l2.b-l1.b)/(l1.k-l2.k),y=l1.k*x+l1.b;
	return l3.k*x+l3.b<y;
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("data.in","r",stdin);
// 	freopen("data.out","w",stdout);
#endif
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		l[i]=Line(1.0/y,1.0/x,i);
	}
	sort(l+1,l+n+1);
	st[++top]=l[1];
	for(int i=2;i<=n;i++)
		if(l[i].k<st[top].k)
		{
			for(;top>1 && check(st[top],st[top-1],l[i]);top--);
			st[++top]=l[i];
		}
	for(int i=1;i<=top;i++)
		ans[++ans[0]]=st[i].id;
	sort(ans+1,ans+ans[0]+1);
	for(int i=1;i<=ans[0];i++)
		printf("%d%c",ans[i],i==ans[0]?'\n':' ');
	return 0;
}
