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;
int n;
struct D {
  int v1;
  int v2;
  vector<int> *i;
  bool operator<(const D &b) const {
    if (v1 != b.v1) return v1 < b.v1;
    return v2 < b.v2;
  }
} inp[1000010];
map<D, int> rev;
vector<int> ans;
vector<int> anp;
long long cross(long long a, long long b, long long c, long long d) {
  return a * d - b * c;
}
int main() {
  scanf("%d", &n);
  int s = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &inp[s].v1, &inp[s].v2);
    if (rev.count(inp[s])) {
      inp[rev[inp[s]]].i->push_back(i);
    } else {
      inp[s].i = new vector<int>();
      inp[s].i->push_back(i);
      rev[inp[s]] = s;
      s++;
    }
  }
  sort(inp, inp + s);
  int maa = -1;
  for (int i = s - 1; i >= 0; i--) {
    if (inp[i].v2 > maa) {
      maa = inp[i].v2;
      if (ans.size() < 2)
        ans.push_back(i);
      else {
        while (ans.size() >= 2) {
          D &p1 = inp[ans.back()], &p2 = inp[ans[ans.size() - 2]];
          if ((long long)inp[i].v1 * p2.v2 * (p2.v1 - p1.v1) *
                      (inp[i].v2 - p1.v2) -
                  (long long)inp[i].v2 * p2.v1 * (p2.v2 - p1.v2) *
                      (inp[i].v1 - p1.v1) <=
              0)
            break;
          ans.pop_back();
        }
        ans.push_back(i);
      }
    }
  }
  for (int i = 0; i < ans.size(); i++) {
    for (int j = 0; j < inp[ans[i]].i->size(); j++)
      anp.push_back((*inp[ans[i]].i)[j]);
  }
  sort(anp.begin(), anp.end());
  for (int i = 0; i < anp.size(); i++) {
    printf("%d%c", anp[i] + 1, " \n"[i == anp.size() - 1]);
  }
  return 0;
}
