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;
pair<pair<long long, long long>, long long> p[800000];
long long maxx, miny, lx, ly;
long long convex[800000], sol[800000], cant;
long long n, i, cont, i1, i2, si;
long long pcruz(pair<pair<long long, long long>, long long> A,
                pair<pair<long long, long long>, long long> B,
                pair<pair<long long, long long>, long long> C) {
  return (B.first.first - A.first.first) * (C.first.second - A.first.second) *
             B.first.second * C.first.first -
         (B.first.second - A.first.second) * (C.first.first - A.first.first) *
             B.first.first * C.first.second;
}
double Vabs(double a) {
  if (a < 0) return -a;
  return a;
}
int main() {
  cin >> n;
  maxx = miny = -1;
  lx = ly = -1;
  for (i = 1; i <= n; i++) {
    scanf("%I64d%I64d", &p[i].first.first, &p[i].first.second);
    if (p[i].first.second >= miny || miny < 0) {
      if (miny == p[i].first.second) {
        if (p[i].first.first > lx || lx < 0) {
          lx = p[i].first.first;
          si = i;
        }
      } else {
        miny = p[i].first.second;
        lx = p[i].first.first;
        si = i;
      }
    }
    p[i].second = i;
  }
  sort(p + 1, p + n + 1);
  reverse(p + 1, p + n + 1);
  convex[1] = n;
  convex[2] = n - 1;
  cont = 2;
  for (i = n - 2; i > 0; i--) {
    while (cont > 1 && pcruz(p[convex[cont - 1]], p[convex[cont]], p[i]) > 0)
      cont--;
    cont++;
    convex[cont] = i;
  }
  cant = 0;
  i = 1;
  while (p[convex[i]].second != si) i++;
  sol[cant++] = p[convex[i]].second;
  while (convex[i] != 1) {
    i++;
    sol[cant++] = p[convex[i]].second;
  }
  sort(sol, sol + cant);
  printf("%I64d", sol[0]);
  for (i = 1; i < cant; i++) printf(" %I64d", sol[i]);
  printf("\n");
  return 0;
}
