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;
const int maxn = (int)2e5 + 10;
map<int, vector<int> > m;
map<int, int> best;
double l[maxn];
double r[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (!best.count(x) || best[x] < y) {
      best[x] = y;
      vector<int> a;
      a.push_back(i + 1);
      m[x] = a;
    } else if (best[x] == y) {
      m[x].push_back(i + 1);
    }
  }
  vector<pair<double, double> > v;
  vector<pair<int, int> > f;
  map<int, int>::iterator it;
  for (it = best.begin(); it != best.end(); it++) {
    v.push_back(make_pair(1.0 / it->first, 1.0 / it->second));
    f.push_back(make_pair(it->first, it->second));
  }
  reverse(v.begin(), v.end());
  for (int i = 0; i < (int)v.size(); i++) {
    r[i] = 1e18;
    double a = v[i].first;
    double b = v[i].second;
    for (int j = 0; j < i; j++) {
      double c = v[j].first;
      double d = v[j].second;
      c -= a;
      d -= b;
      r[i] = min(r[i], -d / c);
    }
    l[i] = 0;
    for (int j = i + 1; j < (int)v.size(); j++) {
      double c = v[j].first;
      double d = v[j].second;
      c -= a;
      d -= b;
      l[i] = max(l[i], d / c);
    }
  }
  vector<int> ans;
  for (int i = 0; i < (int)v.size(); i++) {
    if (l[i] <= r[i] && r[i] > 0) {
      vector<int> a = m[f[i].first];
      for (int j = 0; j < (int)a.size(); j++) {
        ans.push_back(a[j]);
      }
    }
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < (int)ans.size(); i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}
