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 mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
int n, p, q;
map<int, int> nerds;
vector<pair<int, int> > input;
bool addNerd(int p, int q) {
  map<int, int>::iterator iter = nerds.lower_bound(p);
  if (iter == nerds.end()) {
    nerds[p] = q;
    return true;
  } else {
    if (q > iter->second) {
      nerds[p] = q;
      return true;
    }
  }
  return false;
}
void cutNotNerds(int p, int q) {
  map<int, int>::iterator iter = nerds.upper_bound(p);
  map<int, int>::iterator iter1;
  if (iter == nerds.begin()) return;
  iter--;
  while (1) {
    if (iter == nerds.begin()) {
      if (iter->first < p && q >= iter->second) nerds.erase(iter);
      break;
    } else {
      if (p == iter->first) {
        iter--;
        continue;
      }
      if (q < iter->second) break;
      iter1 = --iter;
      nerds.erase(++iter);
      iter = iter1;
    }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < (int)n; i++) {
    cin >> p >> q;
    input.push_back(make_pair(p, q));
    if (addNerd(p, q)) {
      cutNotNerds(p, q);
    }
  }
  vector<pair<int, int> > idxs;
  for (int i = 0; i < (int)n; i++) {
    if (nerds[input[i].first] == input[i].second)
      idxs.push_back(make_pair(input[i].first, i));
  }
  sort(idxs.begin(), idxs.end());
  for (int i = 0; i < (int)idxs.size(); i++) {
    int diffX1 =
        (input[idxs[i].second].first - input[idxs[i - 1].second].first);
    int diffX2 =
        (input[idxs[i + 1].second].first - input[idxs[i].second].first);
    int diffY1 =
        (input[idxs[i - 1].second].second - input[idxs[i].second].second);
    int diffY2 =
        (input[idxs[i].second].second - input[idxs[i + 1].second].second);
    if (i == 0 || idxs.size() - 1)
      cout << idxs[i].second + 1 << " ";
    else if (diffX1 * diffY2 >= diffX2 * diffY1)
      cout << idxs[i].second + 1 << " ";
  }
  cout << endl;
}
