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"

using namespace std;

#define endl '\n'
#define debug(x) cerr << #x << " == " << (x) << '\n';
#define all(x) begin(x), end(x)

using ll = long long;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;

#define x real
#define y imag

template<typename T>
using Segment = array<complex<T>, 2>;

// represents points p with dot(Line.first, p) = Line.second
template<typename T>
using Line = pair<complex<T>, T>;

template<typename T>
T dot(complex<T> a, complex<T> b) { return (conj(a) * b).x(); }

template<typename T>
T cross(complex<T> a, complex<T> b) { return (conj(a) * b).y(); }

template<typename T>
bool cw(complex<T> a, complex<T> b, complex<T> c) { return cross(b - a, c - b) < 0; }

template<typename T>
bool ccw(complex<T> a, complex<T> b, complex<T> c) { return cross(b - a, c - b) > 0; }

template<typename T>
complex<T> normalized(complex<T> a) { return a / gcd<T>(a.x(), a.y()); }

template<typename T>
istream& operator>>(istream& in, complex<T>& p) {
    T x, y;
    in >> x >> y;
    p.x(x), p.y(y);
    return in;
}

template<typename T>
ostream& operator<<(ostream& out, complex<T> p) {
    return out << p.x() << " " << p.y();
}

template<typename T>
bool lex_cmp(complex<T> a, complex<T> b) {
    return a.x() != b.x() ? a.x() < b.x() : a.y() < b.y();
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    using cpx = complex<ll>;

    int n;
    cin >> n;

    vector<cpx> pts(n);
    vector<ll> a(n);

    struct MapCmp {
        bool operator()(cpx a, cpx b) const { return lex_cmp(a, b); }
    };
    map<cpx, int, MapCmp> Id;
    vector<int> I;

    for (int i = 0; i < n; ++i) {
        cin >> pts[i];
        a[i] = pts[i].x() * pts[i].y();
        if (not Id.count(pts[i])) {
            Id[pts[i]] = i;
            I.push_back(i);
        }
    }

    sort(all(I), [&](int i, int j){ return lex_cmp(a[j] * pts[i], a[i] * pts[j]); });

    vector<int> B(n);
    B[0] = I[0];

    int k = 1;
    for (int j = 1; j < n; ++j) {
        int i = I[j];
        while (k > 1 && cw(
                    pts[B[k - 2]] * a[i] * a[B[k - 1]],
                    pts[B[k - 1]] * a[i] * a[B[k - 2]],
                    pts[i] * a[B[k - 1]] * a[B[k - 2]]
                    )) --k;
        if (pts[B[k - 1]].y() > pts[i].y()) B[k++] = i;
    }

    vector<bool> win(n, false);
    for (int i = 0; i < k; ++i) win[B[i]] = true;

    vector<int> ans;
    for (int i = 0; i < n; ++i) {
        if (win[Id[pts[i]]]) ans.push_back(i);
    }

    int r = (int)size(ans);
    for (int i = 0; i < r; ++i) {
        cout << ans[i] + 1 << "\n "[i + 1 < r];
    }

    exit(0);
}
