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 int N = 2e5 + 5;
int n;
pair<pair<int, int>, int> arr[N];
int s[N];
int sz;
vector<int> ans;
set<pair<int, int> > ss;
bool useless(pair<int, int> lft, pair<int, int> mid, pair<int, int> rgt) {
  return 1LL * (lft.second - mid.second) * (rgt.first - mid.first) *
             rgt.second * lft.first >=
         1LL * (mid.second - rgt.second) * (mid.first - lft.first) *
             lft.second * rgt.first;
}
bool lol(pair<int, int> lft, pair<int, int> rgt) {
  long long num = 1LL * (lft.second - rgt.second) * lft.first * rgt.first;
  long long den = 1LL * (rgt.first - lft.first) * lft.second * rgt.second;
  return !num || ((num / llabs(num)) != (den / llabs(den)));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d %d", &arr[i].first.first, &arr[i].first.second);
    arr[i].second = i;
  }
  sort(arr + 1, arr + 1 + n);
  sz = 0;
  for (int i = 1; i <= n; ++i) {
    while (sz >= 1 && arr[s[sz]].first.first == arr[i].first.first &&
           arr[s[sz]].first.second <= arr[i].first.second) {
      --sz;
    }
    while (sz >= 2 &&
           useless(arr[s[sz - 1]].first, arr[s[sz]].first, arr[i].first)) {
      --sz;
    }
    s[++sz] = i;
  }
  for (int i = 1; i <= sz; ++i) {
    if (i < sz) {
      if (lol(arr[s[i]].first, arr[s[i + 1]].first)) {
        continue;
      }
    }
    ss.insert(arr[s[i]].first);
  }
  for (int i = 1; i <= n; ++i) {
    if (ss.find(arr[i].first) != ss.end()) {
      ans.emplace_back(arr[i].second);
    }
  }
  sort(ans.begin(), ans.end());
  for (int x : ans) {
    printf("%d ", x);
  }
}
