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 long long maxn = 2e5 + 500;
const long double eps = 1e-9;
pair<long long, long long> a[maxn];
pair<pair<long long, long long>, long long> b[maxn];
set<pair<long double, long long> > st;
long long nextt[maxn];
bool hazf[maxn];
bool mom[maxn];
bool ans[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long m;
  cin >> m;
  for (long long i = 0; i < m; i++) {
    cin >> a[i].first >> a[i].second;
    b[i].first.first = a[i].first;
    b[i].first.second = a[i].second;
    b[i].second = i;
  }
  sort(a, a + m);
  sort(b, b + m);
  nextt[m - 2] = m - 1;
  for (long long i = m - 2; i > 0; i--) {
    if (a[i].second > a[nextt[i]].second) {
      nextt[i - 1] = i;
    } else {
      hazf[i] = 1;
      nextt[i - 1] = nextt[i];
    }
  }
  {
    long long i = 0;
    if (!(a[i].second > a[nextt[i]].second)) {
      hazf[0] = 1;
    }
  }
  for (long long p = 0; p < m - 1; p++) {
    if (!hazf[p]) {
      st.insert(make_pair((((long double)a[nextt[p]].first - a[p].first) /
                               ((long double)a[p].second - a[nextt[p]].second) -
                           eps * p),
                          p));
    }
  }
  long long maxx = m - 1;
  mom[maxx] = 1;
  for (long long i = 0; i < m - 1; i++) {
    if (a[i].first == a[maxx].first) {
      mom[i] = 1;
    }
  }
  while (st.size() > 0) {
    pair<long long, long long> v = (*(st.begin()));
    st.erase(st.begin());
    if (!hazf[v.second]) {
      long long p = v.second;
      if (nextt[p] == maxx) {
        hazf[maxx] = 1;
        maxx = p;
      } else {
        hazf[nextt[p]] = 1;
        nextt[p] = nextt[nextt[p]];
        st.insert(
            make_pair(((long double)a[nextt[p]].first - a[p].first) /
                              ((long double)a[p].second - a[nextt[p]].second) -
                          eps * p,
                      p));
      }
      mom[maxx] = 1;
    }
  }
  for (long long i = 0; i < m; i++) {
    if (mom[i]) {
      ans[b[i].second] = 1;
    }
  }
  for (long long i = 0; i < m; i++) {
    if (ans[i]) {
      cout << i + 1 << ' ';
    }
  }
}
