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 double EPS = 1e-9;
struct Point {
  double x, y;
  bool operator<(const Point &p) const {
    return x < p.x || (x == p.x && y < p.y);
  }
};
double 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);
}
double det(double aaa, double aab, double aac, double aba, double abb,
           double abc, double aca, double acb, double acc) {
  double ans;
  ans = aaa * ((abb * acc) - (abc * acb)) - aab * ((aba * acc) - (abc * aca));
  ans += aac * ((aba * acb) - (abb * aca));
  return ans;
}
int main() {
  vector<Point> P;
  int n;
  int k = 0, i, t, a, b;
  double pxx = 10e7, pxy = 10e7, pyx = 10e7, pyy = 10e7;
  map<pair<double, double>, int> h;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d%d", &a, &b);
    Point temp;
    temp.x = 1 / (double)a;
    temp.y = 1 / (double)b;
    h[make_pair(temp.x, temp.y)] = i + 1;
    P.push_back(temp);
  }
  vector<Point> H(2 * n);
  sort(P.begin(), P.end());
  for (i = 0; i < n; i++) {
    while (k >= 2 && cross(H[k - 2], H[k - 1], P[i]) < 0) k--;
    H[k++] = P[i];
    if (P[i].x < pxx) {
      pxx = P[i].x;
      pxy = P[i].y;
    }
    if (P[i].x == pxx && P[i].y < pxy) pxy = P[i].y;
    if (P[i].y < pyy) {
      pyx = P[i].x;
      pyy = P[i].y;
    }
  }
  for (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];
    if (P[i].x < pxx) {
      pxx = P[i].x;
      pxy = P[i].y;
    }
    if (P[i].x == pxx && P[i].y < pxy) pxy = P[i].y;
    if (P[i].y < pyy) {
      pyx = P[i].x;
      pyy = P[i].y;
    }
  }
  H.resize(k);
  set<int> sol;
  Point s;
  s.x = pxx;
  s.y = pxy;
  Point r;
  r.x = pyx;
  r.y = pyy;
  for (i = 0; i < (int)H.size(); i++) {
    if (det(1, pxx, pxy, 1, pyx, pyy, 1, H[i].x, H[i].y) <= EPS)
      sol.insert(h[make_pair(H[i].x, H[i].y)]);
  }
  for (set<int>::iterator it = sol.begin(); it != sol.end(); it++)
    printf("%d ", *it);
  puts("");
  return 0;
}
