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 <cstdio>
#include <cstdlib>
#include <cassert>
#include <iostream>
#include <set>
#include <vector>
#include <cstring>
#include <string>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <complex>
#include <map>
#include <queue>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef pair<int, int> pii;
typedef pair<long double, long double> pdd;
typedef vector<pii> vii;
typedef vector<string> vs;

ll p2(ll x) {
    return x*x;
}

bool bad(int a, int b, int c, int d) {
    int d1 = __gcd(a, b);
    int d2 = __gcd(c, d);
    a /= d1; b /= d1;
    c /= d2; d /= d2;
    return (a == c && b == d && d1 < d2);
}

pii pts(vi x) {
    return pdd(x[0], x[1]);
}

pdd sub(pii x, pii y) {
    return pdd((y.first-x.first)/(long double)x.first, (y.second-x.second)/(long double)x.second);
}

long double vect(pdd x, pdd y) {
    return x.first*y.second - x.second*y.first;
}

int n;
vi ok(const vvi & v) {
    vi ok(n);
    vvi q(1, v[0]);
    vi link(n, -1);
    int mx=1,my=1;
    for (int i = 1; i < n; ++i) if (!bad(v[i][0], v[i][1], q.back()[0], q.back()[1])) {
        if (v[i][0] == q.back()[0] && v[i][1] == q.back()[1]) {
            link[i] = q.back()[2];
            continue;
        }
        mx = max(mx, v[i][0]);
        my = max(my, v[i][1]);
        while (!q.empty() && bad(q.back()[0], q.back()[1], v[i][0], v[i][1])) {
            q.pop_back();
        }
        while (q.size() >= 2 && vect(sub(pts(q.back()), pts(v[i])), sub(pts(q[q.size()-2]), pts(v[i]))) > 1e-17) {
            q.pop_back();
        }
        q.push_back(v[i]);
    }
    while (!q.empty() && q.back()[1] < my) q.pop_back();
    while (q.size() >= 2 && q[q.size()-2][1] == my) q.pop_back();
    bool go = 0;
    for (int i = 0; i < q.size(); ++i) {
        if (q[i][0] >= mx && (i + 1 == q.size() || q[i+1][0] < mx)) go = 1;
        if (go) ok[q[i][2]] = 1;
    }
    for (int i = 0; i < n; ++i) if (link[i] >= 0)
        ok[v[i][2]] = ok[link[i]]; 
    return ok;
}

bool cmp(const vi& a, const vi& b) {
    return a[0]*b[1] > a[1]*b[0];
}

int main() {
    cin >> n;
    vvi v(n, vi(3));
    for (int i = 0; i < n; ++i) {
        cin >> v[i][0] >> v[i][1];
        v[i][2] = i;
    }
    sort(v.begin(), v.end(), cmp);
    vi a = ok(v);
    for (int i = 0; i < n; ++i) if (a[i])
        cout << i + 1 << ' ';
    return 0;
}
