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;
long long gcd(long long x, long long y) {
  long long tmp;
  while (y) {
    x %= y;
    tmp = x;
    x = y;
    y = tmp;
  }
  return x;
}
long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
pair<long long, long long> reduce(long long x, long long y) {
  if (y < 0) {
    x = -x;
    y = -y;
  }
  long long g = gcd(abs(x), abs(y));
  x /= g;
  y /= g;
  return make_pair(x, y);
}
pair<long long, long long> multfrac(const pair<long long, long long>& x,
                                    const pair<long long, long long>& y) {
  return reduce(x.first * y.first, x.second * y.second);
}
pair<long long, long long> addfrac(const pair<long long, long long>& x,
                                   const pair<long long, long long>& y) {
  long long l = lcm(abs(x.second), abs(y.second));
  return reduce(l / x.second * x.first + l / y.second * y.first, l);
}
bool fracgt(const pair<long long, long long>& x,
            const pair<long long, long long>& y) {
  return 1.0 * x.first * y.second > 1.0 * x.second * y.first;
}
bool fracge(const pair<long long, long long>& x,
            const pair<long long, long long>& y) {
  return 1.0 * x.first * y.second >= 1.0 * x.second * y.first;
}
pair<long long, long long> intersect(const pair<long long, long long>& i,
                                     const pair<long long, long long>& j) {
  if (i == j) return make_pair(100000, 1);
  return reduce(i.second * j.second * i.first - i.second * j.second * j.first,
                i.first * j.first * j.second - i.first * j.first * i.second);
}
const int MAXN = 200010;
pair<pair<long long, long long>, int> hull[MAXN];
int hsz;
int n;
vector<pair<pair<long long, long long>, int> > sr, tsr;
vector<int> res;
int main() {
  scanf("%d", &n);
  long long s, r;
  for (int i = 0; i < n; i++) {
    cin >> s >> r;
    sr.push_back(make_pair(make_pair(s, r), i + 1));
  }
  sort(sr.begin(), sr.end());
  tsr.push_back(sr[n - 1]);
  long long lr = sr[n - 1].first.second;
  long long ls = sr[n - 1].first.first;
  for (int i = n - 2; i >= 0; i--) {
    if (sr[i].first.second > lr ||
        (sr[i].first.first == ls && sr[i].first.second == lr)) {
      tsr.push_back(sr[i]);
      ls = sr[i].first.first;
      lr = sr[i].first.second;
    }
  }
  sr = tsr;
  int ssz = sr.size();
  hsz = 0;
  hull[hsz++] = sr[0];
  if (sr.size() > 1) hull[hsz++] = sr[1];
  for (int i = 2; i < ssz; i++) {
    while (hsz - 2 >= 0 && fracgt(intersect(sr[i].first, hull[hsz - 2].first),
                                  intersect(sr[i].first, hull[hsz - 1].first)))
      hsz--;
    hull[hsz++] = sr[i];
  }
  for (int i = 0; i < hsz; i++)
    if (hull[i].second > 0) res.push_back(hull[i].second);
  sort(res.begin(), res.end());
  for (int i = 0; i < hsz; i++) printf("%d ", res[i]);
  printf("\n");
  return 0;
}
