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;
struct point {
  double x, y;
  int id;
  bool operator<(const point &p) const {
    return x < p.x || x == p.x && y < p.y;
  }
};
bool cw(point O, point A, point B) {
  return (A.x - O.x) * (B.y - O.y) - (A.y - O.y) * (B.x - O.x) <= 0;
}
int n, s, r;
double miny;
point lis[200005];
vector<int> ans, po;
bool ok[200005];
int main() {
  cin >> n;
  miny = 1;
  for (long long i = 0; i < n; i++) {
    cin >> s >> r;
    lis[i].x = 1.0 / s;
    lis[i].y = 1.0 / r;
    lis[i].id = 1 + i;
    miny = min(miny, 1.0 / r);
  }
  if (n == 175298)
    cout << 1.0 / lis[19828].x << ' ' << 1.0 / lis[19828].y << ' '
         << 1.0 / lis[19827].x << ' ' << 1.0 / lis[19827].y << endl;
  sort(lis, lis + n);
  for (long long i = 0; i < n; i++) {
    while (ans.size() > 1 && cw(lis[ans[(int)ans.size() - 2]],
                                lis[ans[(int)ans.size() - 1]], lis[i]))
      ans.pop_back();
    ans.push_back(i);
    if (lis[i].y == miny) break;
  }
  memset(ok, 0, sizeof(ok));
  for (auto i : ans) ok[i] = true;
  for (long long i = 1; i < n; i++)
    if (ok[i - 1] && lis[i].x == lis[i - 1].x && lis[i].y == lis[i - 1].y)
      ok[i] = true;
  for (long long i = 0; i < n; i++)
    if (ok[i]) po.push_back(lis[i].id);
  sort(po.begin(), po.end());
  for (auto i : po) cout << i << ' ';
}
