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 

//ITNOG
#include<bits/stdc++.h>

using namespace std;

#define int long long
#define rep(i, s, e) for(int i = s; i < e; i ++)
#define x first
#define y second

const int maxN = 200*1000 + 5;
const int mod = 1000*1000*1000 + 9;

pair<pair<int, int>, int> a[maxN];

bool check(pair<int, int> p, pair<int, int> q, pair<int, int> i)
{
    if(p.y == q.y) return false;
    if(i.y == p.y) return false;
    
    return (i.x - p.x) * (p.y - q.y) * q.x * i.y < (q.x - p.x) * (p.y - i.y) * i.x * q.y;
}

main()
{
    ios::sync_with_stdio(0); cin.tie();
    int n; cin >> n;
    rep(i, 0, n) cin >> a[i].x.x >> a[i].x.y, a[i].y = i;
    sort(a, a + n); reverse(a, a + n);

    vector< pair< pair<int, int> , int> > v;
    rep(i, 0, n)
    {
	  if(v.size() && v.back().x.y > a[i].x.y) continue;
	  if(v.size() && v.back().x.y == a[i].x.y && v.back().x.x > a[i].x.x) continue;

//	  cout << a[i].x.x << ' ' << a[i].x.y << endl;

	  while(v.size() > 1 && check(v[v.size() - 2].x, v.back().x, a[i].x)) v.pop_back();
	  v.push_back(a[i]);
    }

    vector<int> ans;
    rep(i, 0, v.size()) ans.push_back(v[i].y + 1);

    sort(ans.begin(), ans.end());
    rep(i, 0, ans.size()) cout << ans[i] << ' ';
    cout << endl;
    return 0;
}
