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>
#include <complex>

#define mp make_pair
#define pb push_back
#define se second
#define fi first
#define M_PI 3.14159265358979323846

typedef long long ll;
typedef long double ld;
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<pair<int, int>, null_type, less<pair<int, int> >, rb_tree_tag, tree_order_statistics_node_update>
typedef complex<long double> base;
const ll NUM = 1e5 + 5;
const ll mod = 1e9 + 7;
const ll INF = 1e18;

struct line{
    int k, b;
    ld start;
    line(int k, int b){
        this->k = k;
        this->b = b;
    }
    line(){
    }
};

ld inter(line a, line b){
    return (ld) (1 / (ld) a.b - 1 / (ld) b.b) / (ld) (1 / (ld) b.k - 1 / (ld) a.k);
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    set<pair<int, int> > st;
    vector<pair<pair<int, int>, int> > cur;
    vector<pair<int, int> > res;
    for (int i = 0; i < n; i++){
        int s, r;
        cin >> s >> r;
        cur.pb({{r, s}, i});
    }
    sort(cur.begin(), cur.end());
    reverse(cur.begin(), cur.end());
    for (int i = 0; i < n; i++){
        int p = i;
        while (p + 1 < n && cur[p + 1].fi.fi == cur[i].fi.fi)
            p++;
        res.pb(cur[i].fi);
        i = p;
    }
    vector<line> cht;
    for (int i = 0; i < res.size(); i++){
        line l(res[i].fi, res[i].se);
        while(!cht.empty() && inter(l, cht[cht.size() - 1]) > cht[cht.size() - 1].start)
            cht.erase(cht.begin() + cht.size() - 1);
        if (!cht.empty())
            l.start = inter(l, cht[cht.size() - 1]);
        else
            l.start = INF;
        cht.pb(l);
    }
    for (auto l : cht)
        st.insert({l.k, l.b});
    vector<int> ans;
    for (int i = 0; i < cur.size(); i++)
        if (st.find({cur[i].fi.fi, cur[i].fi.se}) != st.end())
            ans.pb(cur[i].se);
    sort(ans.begin(), ans.end());
    for (int i = 0; i < ans.size(); i++)
        cout << ans[i] + 1 << ' ';
    return 0;
}