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;
struct pt {
  long double x, y;
  basic_string<int> id;
  bool operator<(pt b) const {
    if (x != b.x) return x < b.x;
    return y < b.y;
  }
};
bool ispod(pt a, pt b) { return a.x <= b.x && a.y <= b.y; }
long double det(pt a, pt b, pt c) {
  return a.x * b.y + b.x * c.y + c.x * a.y - a.y * b.x - b.y * c.x - c.y * a.x;
}
int n, nn, m, h;
pt a[200005], b[200005], c[200005];
map<pair<int, int>, int> mp;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x, y;
    cin >> x >> y;
    if (mp.count({x, y}))
      a[mp[{x, y}]].id += i;
    else {
      a[nn].x = (long double)1 / x;
      a[nn].y = (long double)1 / y;
      a[nn].id = {i};
      mp[{x, y}] = nn;
      nn++;
    }
  }
  sort(a, a + nn);
  for (int i = 0; i < nn; i++) {
    while (h >= 2 && det(c[h - 2], c[h - 1], a[i]) < -1e-16) h--;
    c[h++] = a[i];
  }
  {
    basic_string<int> sol;
    for (int i = 0; i < h; i++) sol += c[i].id;
    sort(sol.begin(), sol.end());
    for (int x : sol) cerr << x << ' ';
    cerr << '\n';
  }
  while (h >= 2 && ispod(c[h - 2], c[h - 1])) h--;
  basic_string<int> sol;
  for (int i = 0; i < h; i++) sol += c[i].id;
  sort(sol.begin(), sol.end());
  for (int x : sol) cout << x << ' ';
  cout << '\n';
}
