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 N = 200010;
vector<int> v[N];
map<pair<int, int>, int> mp;
pair<int, int> pi[N];
bool check(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
  return 1LL * (b.first - a.first) * a.second * c.first *
             (b.second - c.second) <
         1LL * (c.first - b.first) * c.second * a.first * (a.second - b.second);
}
int main() {
  int n;
  scanf("%d", &n);
  int tot = 0;
  for (int i = 1; i <= n; i++) {
    pair<int, int> a;
    scanf("%d%d", &a.first, &a.second);
    if (!mp.count(a)) {
      mp[a] = ++tot;
      pi[tot] = a;
    }
    int id = mp[a];
    v[id].push_back(i);
  }
  sort(pi + 1, pi + tot + 1);
  int i = 1;
  int cnt = 0;
  while (i <= tot) {
    int j = i;
    while (j <= tot && pi[i].first == pi[j].first) j++;
    pi[++cnt] = pi[j - 1];
    i = j;
  }
  int cc = 1;
  for (int i = 2; i <= tot; i++) {
    while (cc && pi[cc].second <= pi[i].second) cc--;
    pi[++cc] = pi[i];
  }
  stack<pair<int, int> > stk;
  stk.push(pi[1]);
  for (int i = 2; i <= cc; i++) {
    while (stk.size() > 1) {
      pair<int, int> b = stk.top();
      stk.pop();
      pair<int, int> a = stk.top();
      stk.pop();
      if (check(a, b, pi[i])) {
        stk.push(a);
      } else {
        stk.push(a);
        stk.push(b);
        break;
      }
    }
    stk.push(pi[i]);
  }
  vector<int> ans;
  while (!stk.empty()) {
    int id = mp[stk.top()];
    stk.pop();
    for (int i = 0; i < v[id].size(); i++) ans.push_back(v[id][i]);
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}
