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;
const double EPS = 1e-7;
const int MOD = (int)1e9 + 7;
const int MAXN = (int)1e5 + 10;
struct point {
  long long x, y;
  point() { x = y = 0; }
  point(long long X, long long Y) { x = X, y = Y; }
  pair<long long, long long> pair() { return make_pair(x, y); }
  void print() { cout << "<" << x << ";" << y << ">" << '\n'; }
};
long long cross(point a, point b, point c) {
  return b.x * a.y * (c.y - b.y) * (c.x - a.x) +
         b.y * a.x * (c.x - b.x) * (a.y - c.y);
}
map<pair<long long, long long>, vector<int> > M;
bool compy(point a, point b) {
  return make_pair(a.y, a.x) > make_pair(b.y, b.x);
}
bool compx(point a, point b) {
  return make_pair(a.x, a.y) > make_pair(b.x, b.y);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<point> V;
  for (int i = 0; i < n; ++i) {
    long long a, b;
    cin >> a >> b;
    M[make_pair(a, b)].push_back(i + 1);
    V.push_back(point(a, b));
  }
  if ((int)M.size() == 1) {
    for (int i = 1; i <= n; ++i) cout << i << " ";
    cout << '\n';
    return 0;
  }
  point min_y = *min_element(V.begin(), V.end(), compy);
  point min_x = *min_element(V.begin(), V.end(), compx);
  if (min_y.pair() == min_x.pair()) {
    vector<int> &v = M[make_pair(min_y.x, min_y.y)];
    for (int i = 0; i < (int)v.size(); ++i) cout << v[i] << " ";
    cout << '\n';
    return 0;
  }
  sort(V.begin(), V.end(), compx);
  vector<point> cHull(n);
  int t = 0;
  for (int i = 0; !i || V[i - 1].pair() != min_y.pair(); cHull[t++] = V[i++])
    while (t >= 2 && cross(cHull[t - 2], cHull[t - 1], V[i]) <= 0) --t;
  vector<int> ans;
  for (int i = 0; i < t; ++i) {
    vector<int> &cur = M[cHull[i].pair()];
    for (int j = 0; j < (int)cur.size(); ++j) {
      ans.push_back(cur[j]);
    }
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < (int)ans.size(); ++i) cout << ans[i] << " ";
  cout << '\n';
  return 0;
}
