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<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#define F first
#define S second
#define ll long long
#define ull  unsigned long long
#define pb push_back

using namespace std;
using namespace __gnu_pbds;
typedef  tree< int , null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
const int N = 500000;
double from[N];
int p[N], k[N], b[N];
pair < pair <double, double >, int > a[N];
main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++){
        cin >> a[i].F.F >> a[i].F.S;
        //a[i].F.F = rand() % 32 + 1;
        //a[i].F.S = rand() % 32431 + 1;
        a[i].F.F = 1 / a[i].F.F;
        a[i].F.S = 1 / a[i].F.S;
        a[i].S = i;
    }
    sort(a + 1, a + n + 1);
    reverse(a + 1, a + n + 1);
    int sz = 1;
    from[0] = -1e9;
    k[0] = a[1].F.F;
    b[0] = a[1].F.S;
    p[0] = a[1].S;
    for (int i = 2; i <= n; i++){
        double ki = a[i].F.F;
        double bi = a[i].F.S;
        while(1){
            double K = k[sz - 1];
            double B = b[sz - 1];
            double x = (bi - B) / (K - ki);
            if (abs(k[sz - 1] - ki) < 1e-8){
                while(sz > 0 && b[sz - 1] > bi && abs(k[sz - 1] - ki) < 1e-8) sz--;
                if (abs(k[sz - 1] - ki) < 1e-8 || sz == 0){
                    k[sz] = ki;
                    b[sz] = bi;
                    p[sz] = a[i].S;
                    from[sz] = x;
                    sz++;
                    break;
                }
            }
            if (from[sz - 1] - 1e-6 < x){
                k[sz] = ki;
                b[sz] = bi;
                p[sz] = a[i].S;
                from[sz] = x;
                sz++;
                break;
            } else sz--;
        }
    }
    vector < int > ans;
    for (int i = 0; i < sz; i++) ans.pb(p[i]);
    sort(ans.begin(), ans.end());
    for (auto i: ans) cout << i << " ";
}
