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;
const long double EPS = 1e-24;
const int N = 2e5 + 2;
struct pt {
  long double x, y;
  vector<int> id;
  pt(long double a = 0, long double b = 0) : x(a), y(b) {}
  pt operator-(const pt &p) const { return pt(x - p.x, y - p.y); }
  bool operator<(const pt &p) const {
    return x < p.x || (abs(x - p.x) < EPS && y < p.y);
  }
};
long double area(const pt &a, const pt &b) { return a.x * b.y - a.y * b.x; }
long double d(const pt &p) { return p.x * p.x + p.y * p.y; }
pair<pair<int, int>, int> in[N];
pt arr[N], que[N], base(2, 2);
bool cmp(const pt &a, const pt &b) {
  return area(a - base, b - base) < -EPS ||
         (abs(area(a - base, b - base)) < EPS && d(a - base) > d(b - base));
}
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i)
    cin >> in[i].first.first >> in[i].first.second, in[i].second = i;
  sort(in + 1, in + 1 + n);
  int m = 0;
  for (int i = 1; i <= n; ++i) {
    if (in[i].first != in[i - 1].first)
      arr[++m] = pt(1.0 / in[i].first.first, 1.0 / in[i].first.second);
    arr[m].id.push_back(in[i].second);
  }
  sort(arr + 1, arr + m + 1, cmp);
  que[1] = arr[1];
  int r = 1;
  for (int i = 2; i <= m; ++i) {
    while (r >= 2 && area(que[r] - que[r - 1], arr[i] - que[r]) > 0) --r;
    que[++r] = arr[i];
  }
  vector<int> ans;
  for (int i = 1; i <= r; ++i) {
    if ((i == r || que[i + 1].y > que[i].y + EPS) &&
        (i == 1 || que[i].x + EPS < que[i - 1].x))
      for (auto it : que[i].id) ans.push_back(it);
  }
  sort(ans.begin(), ans.end());
  for (auto it : ans) cout << it << ' ';
  return 0;
}
