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;
double ab(double a) { return a < 0 ? -a : a; }
struct pt;
vector<pt> pepr;
const double EPS = 0.0000000001;
struct pt {
  double x, y;
  long long id;
  pt(double a = 0, double b = 0, long long i = 0) {
    x = a;
    y = b;
    id = i;
  }
  pt operator-(const pt cp) const { return pt(x - cp.x, y - cp.y); }
  long long operator^(const pt cp) const { return x * cp.y - y * cp.x; }
  bool operator<(const pt cp) const {
    long long cr = ((*this - pepr[0]) ^ (cp - pepr[0]));
    if (ab(cr) > EPS) return cr > 0;
    return x < cp.x;
  }
};
vector<pt> ee;
void wrap() {
  long long mid = 0;
  for (long long i = 0; i < ((long long((pepr).size()))); ++i) {
    if (pepr[i].x < pepr[mid].x ||
        (pepr[i].x == pepr[mid].x && pepr[i].y < pepr[mid].y))
      mid = i;
  }
  if (mid != 0) swap(pepr[0], pepr[mid]);
  sort(pepr.begin() + 1, pepr.end());
  ee.push_back(pepr[0]);
  pt ppr, pr, cr;
  for (long long i = 1; i < (long long((pepr).size())); ++i) {
    while (1) {
      if ((long long((ee).size())) == 1) break;
      ppr = ee[(long long((ee).size())) - 2];
      pr = ee[(long long((ee).size())) - 1];
      cr = pepr[i];
      if (((pr - ppr) ^ (cr - pr)) >= -EPS) break;
      ee.pop_back();
    }
    ee.push_back(pepr[i]);
  }
  while (1) {
    if ((long long((ee).size())) <= 1) break;
    pr = ee[(long long((ee).size())) - 2];
    cr = ee[(long long((ee).size())) - 1];
    if (cr.y < ee[0].y && cr.y < pr.y) break;
    ee.pop_back();
  }
}
set<pair<long long, long long> > opst;
long long n;
map<long long, vector<long long> > mx, my;
vector<long long> px, py;
vector<long long> eli;
set<pair<long long, long long> > isok;
vector<long long> ops;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  long long r, s;
  eli.resize(n, 0);
  for (long long i = 0; i < (n); ++i) {
    cin >> r >> s;
    px.push_back(r);
    py.push_back(s);
    mx[r].push_back(i);
    my[s].push_back(i);
  }
  for (auto i = (mx).begin(); i != (mx).end(); ++i) {
    long long mni = i->second[0];
    for (long long j = 0; j < ((long long((i->second).size()))); ++j) {
      if (py[i->second[j]] > py[mni]) mni = i->second[j];
    }
    for (long long j = 0; j < ((long long((i->second).size()))); ++j) {
      if (py[i->second[j]] != py[mni]) eli[i->second[j]] = 1;
    }
  }
  for (auto i = (my).begin(); i != (my).end(); ++i) {
    long long mni = i->second[0];
    for (long long j = 0; j < ((long long((i->second).size()))); ++j) {
      if (px[i->second[j]] > px[mni]) mni = i->second[j];
    }
    for (long long j = 0; j < ((long long((i->second).size()))); ++j) {
      if (px[i->second[j]] != px[mni]) eli[i->second[j]] = 1;
    }
  }
  for (long long i = 0; i < ((long long((px).size()))); ++i) {
    if (eli[i]) continue;
    if (opst.find(make_pair(px[i], py[i])) != opst.end()) continue;
    opst.insert(make_pair(px[i], py[i]));
    pepr.push_back(pt(1.0 / px[i], 1.0 / py[i], i));
  }
  wrap();
  for (long long i = 0; i < ((long long((ee).size()))); ++i) {
    isok.insert(make_pair(px[ee[i].id], py[ee[i].id]));
  }
  for (long long i = 0; i < ((long long((px).size()))); ++i) {
    if (eli[i]) continue;
    if (isok.find(make_pair(px[i], py[i])) != isok.end()) ops.push_back(i + 1);
  }
  for (long long i = 0; i < ((long long((ops).size()))); ++i)
    cout << ops[i] << " ";
  cout << "\n";
  return 0;
}
