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;
pair<long long, long long> o, oo, stk[20010];
vector<pair<long long, long long> > pnt;
vector<int> ans;
map<pair<long long, long long>, vector<int> > mp;
int tt, top, n;
long long xx, yy;
bool cmp1(pair<long long, long long> i, pair<long long, long long> j) {
  if (i.second != j.second) return i.second > j.second;
  return i.first > j.first;
}
bool cmp2(pair<long long, long long> i, pair<long long, long long> j) {
  return i > j;
}
long long cj(pair<long long, long long> i, pair<long long, long long> j,
             pair<long long, long long> k) {
  return (j.second * k.first * i.first * i.second -
          j.second * i.first * k.first * k.second -
          i.second * k.first * j.first * j.second +
          j.first * k.first * j.second * k.second) -
         (j.first * k.second * i.first * i.second -
          j.first * i.second * k.first * k.second -
          i.first * k.second * j.first * j.second +
          j.first * k.first * j.second * k.second);
}
long long sqr(long long xx) { return xx * xx; }
bool cmp3(pair<long long, long long> i, pair<long long, long long> j) {
  long long tmp = cj(o, i, j);
  pair<long long, long long> t1 = make_pair(i.first - o.first,
                                            i.second - o.second),
                             t2 = make_pair(j.first - o.first,
                                            j.second - o.second);
  if (tmp == 0)
    return sqr(t1.first) + sqr(t1.second) < sqr(t2.first) + sqr(t2.second);
  return tmp < 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d%I64d", &xx, &yy);
    mp[make_pair(xx, yy)].push_back(i);
  }
  if (mp.size() == 1) {
    for (int i = 1; i <= n; i++) printf("%d ", i);
    exit(0);
  }
  for (map<pair<long long, long long>, vector<int> >::iterator p = mp.begin();
       p != mp.end(); p++)
    pnt.push_back(make_pair((p->first).first, (p->first).second));
  sort(pnt.begin(), pnt.end(), cmp1);
  o = pnt[0];
  sort(pnt.begin(), pnt.end(), cmp2);
  oo = pnt[0];
  if (o == oo) {
    for (vector<int>::iterator p = mp[o].begin(); p != mp[o].end(); p++)
      printf("%d ", *p);
    exit(0);
  }
  sort(pnt.begin(), pnt.end(), cmp3);
  stk[++top] = pnt[0];
  stk[++top] = pnt[1];
  tt = top;
  while (tt < pnt.size() && stk[top] != oo) {
    while (top > 1 && cj(stk[top - 1], stk[top], pnt[tt]) > 0) top--;
    stk[++top] = pnt[tt++];
  }
  for (int i = 1; i <= top; i++)
    for (vector<int>::iterator p = mp[stk[i]].begin(); p != mp[stk[i]].end();
         p++)
      ans.push_back(*p);
  sort(ans.begin(), ans.end());
  for (int i = 0, ed = ans.size(); i < ed; i++) printf("%d ", ans[i]);
  return 0;
}
