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 mp make_pair
#define X first
#define Y second

#define int long long

typedef pair<pair<int, int>, int> ppdi;

const int maxn = 200 * 1000 + 5, mod = 1000*1000*1000 + 7;

int n;;

bool is_good(ppdi a, ppdi b, ppdi c)
{
    if(a.X.Y == b.X.Y || c.X.Y == a.X.Y)
        return false;
    return ((c.X.X - a.X.X) * (a.X.Y - b.X.Y) * b.X.X * c.X.X < (b.X.X - a.X.X) * (a.X.Y - c.X.Y) * c.X.X * b.X.Y);

}


ppdi a[maxn];

main()
{
    ios_base::sync_with_stdio(false);cin.tie();
    cin>>n;
    for(int i = 0; i < n; i++)
        cin>>a[i].X.X>>a[i].X.Y, a[i].Y = i;
    sort(a, a + n);
    reverse(a, a + n);
    vector <ppdi> cros;
    int maxim = -1;
    for(int i = 0; i < n; i++)
    {
        if(maxim >= a[i].X.Y)
            continue;
        maxim = a[i].X.Y;
        while(cros.size() > 1 && is_good(cros[cros.size() - 2], cros.back(), a[i]))
            cros.pop_back();
        cros.push_back(a[i]);
    }

    set<pair< int, int> > sag;
    vector <int> ans;
    for(auto &i : cros)
        sag.insert(i.X);
    for(int i = 0; i < n; i++)
        if(sag.find(a[i].X) != sag.end())
            ans.push_back(a[i].Y + 1);
    sort(ans.begin(), ans.end());
    for(auto &i : ans)
        cout<<i<<" ";





    return 0;
}
