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;
typedef long long coord_t;
typedef long long coord2_t;
struct Point {
  coord_t x, y, ct;
  bool operator<(const Point &p) const {
    return x < p.x || (x == p.x && y < p.y);
  }
};
coord2_t cross(const Point &O, const Point &A, const Point &B) {
  return (long)(A.x - O.x) * (B.y - O.y) - (long)(A.y - O.y) * (B.x - O.x);
}
vector<Point> convex_hull(vector<Point> P) {
  int n = P.size(), k = 0;
  vector<Point> H(2 * n);
  sort(P.begin(), P.end());
  for (int i = 0; i < n; ++i) {
    while (k >= 2 && cross(H[k - 2], H[k - 1], P[i]) < 0) k--;
    H[k++] = P[i];
  }
  for (int i = n - 2, t = k + 1; i >= 0; i--) {
    while (k >= t && cross(H[k - 2], H[k - 1], P[i]) < 0) k--;
    H[k++] = P[i];
  }
  H.resize(k);
  return H;
}
int main() {
  int n;
  cin >> n;
  vector<Point> ch;
  ch.push_back({0LL, 0LL, 0LL});
  for (long long ctr1 = 0; ctr1 < n; ctr1++) {
    long long a, b;
    cin >> a >> b;
    ch.push_back({a, b, ctr1 + 1});
  }
  vector<Point> rez = convex_hull(ch);
  for (int ctr1 = 0; ctr1 < rez.size(); ctr1++) {
  }
  vector<int> r;
  for (int ctr1 = 0; ctr1 < rez.size(); ctr1++) r.push_back(rez[ctr1].ct);
  sort(r.begin(), r.end());
  for (int ctr1 = 0; ctr1 < r.size(); ctr1++)
    if (r[ctr1] != 0) cout << r[ctr1] << " ";
  return 0;
}
