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 

//God & me
//be konkoor salami dobare bayad kard ;)
#include <bits/stdc++.h>
#define pb push_back
#define X first
#define Y second
#define int long long
using namespace std;
template <class T, class L> bool smax(T &x,L y){  return x < y ? (x = y, 1) : 0;  }
template <class T, class L> bool smin(T &x,L y){  return y < x ? (x = y, 1) : 0;  }
typedef pair<int, int> pii;

const int maxn = 2e5 + 17;
int n;
struct S{int x, y, id;} a[maxn];
int cross(pii a, pii b){
    return a.X * b.Y - a.Y * b.X;
}
int cross(S a, S b, S c){
    return cross({b.x - a.x, b.y - a.y}, {c.x - a.x, c.y - a.y});
}
bool check(S a, S b, S c){
    int mx = a.x * b.x * c.x, my = a.y * b.y * c.y;
    a.x = mx / a.x, a.y = my / a.y;
    b.x = mx / b.x, b.y = my / b.y;
    c.x = mx / c.x, c.y = my / c.y;
    return cross(c, b, a) < 0;
}
main(){
    ios::sync_with_stdio(0), cin.tie(0);
    cin >> n;
    for(int i = 0; i < n; i++)
	cin >> a[i].x >> a[i].y, a[i].id = i;
    sort(a, a + n, [](S &a, S &b){  return pii(a.x, a.y) < pii(b.x, b.y);  });
    vector<int>s;
    for(int i = 0; i < n; i++){
	while(s.size() && a[s.back()].y <= a[i].y && pii(a[s.back()].x, a[s.back()].y) != pii(a[i].x, a[i].y))  s.pop_back();
	while(s.size() > 1 && check(a[ s[s.size() - 2] ], a[s.back()], a[i]))  s.pop_back();
	s.pb(i);
    }
    sort(s.begin(), s.end(), [](int i, int j){  return a[i].id < a[j].id;  });
    for(auto x : s)
	cout << a[x].id + 1 << ' ';
    cout << '\n';
    return 0;
}
