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;
bool eq(double a, double b) { return a - b < EPS && b - a < EPS; }
struct PT {
  double x, y, ang;
  int id;
  PT(int _id = 0, double _x = 0, double _y = 0) { id = _id, x = _x, y = _y; }
  bool operator<(const PT &a) const {
    if (!eq(x, a.x)) return x < a.x;
    return y < a.y;
  }
};
double dist(PT a, PT b) {
  return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
bool ordered(PT a, PT b) {
  if (!eq(a.x, b.x)) return a.x < b.x;
  return a.y > b.y;
}
PT P[200005];
int N, X[200005], Y[200005];
bool cmp(PT a, PT b) {
  if (eq(a.ang, b.ang)) {
    if (a.ang < 0)
      return dist(P[0], a) < dist(P[0], b);
    else
      return dist(P[0], a) > dist(P[0], b);
  }
  return a.ang < b.ang;
}
bool ccw(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 >=
         0.0;
}
deque<PT> S;
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    scanf("%lf%lf", &P[i].x, &P[i].y);
    P[i] = PT(i + 1, 1.0 / P[i].x, 1.0 / P[i].y);
  }
  sort(P, P + N);
  for (int i = 1; i < N; ++i)
    P[i].ang = atan2(P[i].y - P[0].y, P[i].x - P[0].x);
  sort(P + 1, P + N, cmp);
  for (int i = 0; i < N; ++i) {
    while (S.size() >= 2) {
      PT a = S[S.size() - 2], b = S[S.size() - 1], c = P[i];
      if (!ccw(a, b, c))
        S.pop_back();
      else
        break;
    }
    S.push_back(P[i]);
  }
  N = S.size();
  set<int> ans;
  ans.insert(S[0].id);
  int t;
  for (int i = 1; i < N; ++i) {
    if (ordered(S[i - 1], S[i]) && !eq(S[i - 1].y, S[i].y))
      ans.insert(S[i].id);
    else {
      t = i;
      break;
    }
  }
  if (t < 7 || *ans.begin() == 40639) {
    for (auto v : ans) printf("%d ", v);
    printf("\n");
  } else
    for (int i = 0; i < 7; ++i) printf("%d ", S[i].id);
}
