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;
double f(pair<int, int> &i, pair<int, int> &j) {
  double s_i = i.first;
  double r_i = i.second;
  double s_j = j.first;
  double r_j = j.second;
  return ((1 / r_j) - (1 / r_i)) / ((1 / s_i) - (1 / s_j));
}
double findMaximum(vector<pair<int, int> > &points, pair<int, int> origin) {
  int l = -1;
  int r = points.size() - 1;
  while (r - l > 1) {
    int mid = (l + r) >> 1;
    if (f(points[mid], origin) > f(points[mid + 1], origin))
      r = mid;
    else
      l = mid;
  }
  return f(points[l + 1], origin);
}
double findMinimum(vector<pair<int, int> > &points, pair<int, int> origin) {
  int l = -1;
  int r = points.size() - 1;
  while (r - l > 1) {
    int mid = (l + r) >> 1;
    if (f(points[mid], origin) < f(points[mid + 1], origin))
      r = mid;
    else
      l = mid;
  }
  return f(points[l + 1], origin);
}
double cross(pair<int, int> o, pair<int, int> a, pair<int, int> b) {
  return (a.first - o.first) * (b.second - o.second) -
         (b.first - o.first) * (a.second - o.second);
}
vector<int> solve(vector<int> &S, vector<int> &R) {
  vector<pair<int, int> > points;
  int n = S.size();
  for (int i = 0; i < n; ++i) points.push_back({S[i], R[i]});
  sort(points.begin(), points.end());
  points.erase(unique(points.begin(), points.end()), points.end());
  vector<pair<int, int> > candidates;
  for (int i = points.size() - 1; i >= 0; --i)
    if (candidates.empty() || points[i].second > candidates.back().second)
      candidates.push_back(points[i]);
  reverse(candidates.begin(), candidates.end());
  set<pair<int, int> > goodPairs;
  goodPairs.insert(candidates[0]);
  goodPairs.insert(candidates.back());
  map<pair<int, int>, double> maxTime, minTime;
  vector<pair<int, int> > Hull;
  Hull.push_back(candidates[0]);
  for (int i = 1; i < candidates.size() - 1; ++i) {
    minTime[candidates[i]] = findMinimum(Hull, candidates[i]);
    Hull.push_back(candidates[i]);
  }
  Hull.clear();
  Hull.push_back(candidates.back());
  for (int i = candidates.size() - 2; i > 0; --i) {
    maxTime[candidates[i]] = findMaximum(Hull, candidates[i]);
    Hull.push_back(candidates[i]);
  }
  for (int i = 1; i < candidates.size() - 1; ++i)
    if (minTime[candidates[i]] <= maxTime[candidates[i]])
      goodPairs.insert(candidates[i]);
  vector<int> ans;
  for (int i = 0; i < n; ++i)
    if (goodPairs.find({S[i], R[i]}) != goodPairs.end()) ans.push_back(i);
  return ans;
}
int main() {
  cin.sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> S, R;
  for (int i = 0; i < n; ++i) {
    int s, r;
    cin >> s >> r;
    S.push_back(s);
    R.push_back(r);
  }
  vector<int> ans = solve(S, R);
  for (auto i : ans) cout << i + 1 << " ";
  return 0;
}
