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 <fstream>
#include <string.h>
#include <memory.h>
#include <stdio.h>
#include <algorithm>
#include <list>
#include <queue>
#include <string>
#include <math.h>
#include <vector>
#include <set>
#include <map>
using namespace std;
struct Freg{
	long long y, x;
	bool operator <(Freg &t){
		return y*t.x < x*t.y;
	}
};
struct  Node
{
	int u;
	Freg dif;
};
int N;
int a[200005];
int b[200005];
int index[200005];
list <Node> ll;
int cmp(int x, int y){
	if (x != y)
		return a[x]>a[y];
	else
		return b[x] > b[y];
}
int cmp2;
Freg diff(int former, int latter){
	Freg res;
	res.y = (long long)b[former] * b[latter] * (a[former] - a[latter]);
	res.x = (long long)a[former] * a[latter] * (b[latter] - b[former]);
	return res;
}
int main()
{
	int i, j;
	int u;
	int lastb;
	Node temp;
	Node t;
	cin >> N;
	for (i = 0; i < N; ++i){
		scanf("%d%d", &a[i], &b[i]);
		index[i] = i;
	}
	sort(index, index + N, cmp);
	temp.u = index[0];
	temp.dif.y = 0;
	temp.dif.x = 1;
	ll.push_back(temp);
	lastb = b[index[0]];
	for (i = 1; i < N; ++i){
		u = index[i];
		if (b[u] <= lastb)
			continue;
		lastb = b[u];
		while (1){
			t = ll.back();
			temp.dif = diff(t.u, u);
			if (t.dif < temp.dif){
				temp.u = u;
				ll.push_back(temp);
				break;
			}
			else{
				ll.pop_back();
			}
		}
	}
	for (i = 0; i < 200000; ++i){
		ll.push_back(temp);
	}
	list  <Node>::iterator it, end;
	it = ll.begin();
	end = ll.end();
	i = 0;
	N = ll.size();
	for (; it != end; ++it){
		index[i++] = it->u;
	}
	sort(index, index + N);
	printf("%d", index[0]+1);
	for (i = 1; i < N; ++i)
		printf(" %d", index[i]+1);
	printf("\n");

	return 0;
}