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;
pair<pair<int, int>, int> pos[200010];
pair<double, double> co[200010];
int N;
int stack[200010];
int S;
double intx(pair<double, double> &a, pair<double, double> &b) {
  double x = (b.second - a.second) / (a.first - b.first);
  return x;
}
int comp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.first.first != b.first.first ? a.first.first < b.first.first
                                        : a.first.second > b.first.second;
}
vector<int> inc[200010];
int st;
void init() {
  sort(pos, pos + N, comp);
  int i;
  double x1, x2;
  for (i = 0; i < N; ++i) {
    co[i] = make_pair(1.0 / pos[i].first.first, 1.0 / pos[i].first.second);
  }
  for (i = 0; i < N; ++i) {
    if (i && pos[i - 1].first.first == pos[i].first.first) {
      if (pos[i - 1].first.second > pos[i].first.second) {
        continue;
      }
      inc[i - 1].push_back(i);
      continue;
    }
    while (S > 1) {
      x1 = intx(co[i], co[stack[S - 2]]);
      x2 = intx(co[stack[S - 1]], co[stack[S - 2]]);
      if (x2 - x1 <= (1e-7)) break;
      --S;
    }
    stack[S++] = i;
  }
  for (i = 1; i < S; ++i) {
    x1 = intx(co[stack[i]], co[stack[i - 1]]);
    if (x1 > -(1e-7)) break;
  }
  st = i - 1;
}
int comp1(int i, int j) { return pos[i].second < pos[j].second; }
int main() {
  ios::sync_with_stdio(0);
  cin >> N;
  int i;
  for (i = 0; i < N; ++i) {
    cin >> pos[i].first.first >> pos[i].first.second;
    pos[i].second = i + 1;
  }
  init();
  for (i = st; i < S; ++i) {
    for (int k = 0; k < inc[stack[i]].size(); ++k)
      stack[S++] = inc[stack[i]][k];
  }
  sort(stack + st, stack + S, comp1);
  for (i = st; i < S; ++i) {
    cout << pos[stack[i]].second << " ";
  }
  return 0;
}
