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>
using namespace std;
map<int, int> mp;
pair<pair<double, double>, int> a[200005];
pair<int, int> b[200005];
vector<pair<pair<double, double>, int> > v;
bool comp(pair<pair<double, double>, int> a,
          pair<pair<double, double>, int> b) {
  if (a.first.first > b.first.first)
    return true;
  else if (a.first.first < b.first.first)
    return false;
  return a.first.second < b.first.second;
}
long long fun(pair<pair<double, double>, int> p1,
              pair<pair<double, double>, int> p2,
              pair<pair<double, double>, int> p3) {
  double d1 = (double)(p3.first.second - p2.first.second) /
              (p2.first.first - p3.first.first);
  double d2 = (double)(p1.first.second - p2.first.second) /
              (p2.first.first - p1.first.first);
  return d1 <= d2;
}
long long ans[200005];
int main() {
  long long i, j, k, l, m, n, t;
  cin >> n;
  for (i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    b[i].first = x;
    b[i].second = y;
    a[i].first.first = (double)1 / x;
    a[i].first.second = (double)1 / y;
    a[i].second = i;
  }
  sort(a, a + n, comp);
  for (i = 0; i < n; i++) {
    if (i > 0 && a[i - 1].first.first - a[i].first.first <= 0.0000000000001) {
      continue;
    }
    while (v.size() >= 2 && fun(v[v.size() - 2], v[v.size() - 1], a[i]))
      v.pop_back();
    v.push_back(a[i]);
  }
  double val = 1000000000;
  int pos;
  for (i = 0; i < v.size(); i++) {
    double temp = v[i].first.second;
    if (temp <= val) {
      pos = i;
      val = temp;
    }
  }
  for (i = 0; i < n; i++) {
    ans[i] = 0;
  }
  for (i = pos; i < v.size(); i++) {
    ans[v[i].second] = 1;
  }
  for (i = 0; i < n; i++) {
    if (ans[i] == 1) {
      mp.insert(make_pair(b[i].first, b[i].second));
    }
  }
  for (i = 0; i < n; i++) {
    if (ans[i] == 0 && mp.find(b[i].first) != mp.end() &&
        mp[b[i].first] == b[i].second) {
      ans[i] = 1;
    }
  }
  for (i = 0; i < v.size(); i++) {
    ans[v[i].second] = 1;
  }
  for (i = 0; i < n; i++) {
    if (ans[i] == 1) {
      printf("%d ", i + 1);
    }
  }
  return 0;
}
