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;
long long r[200002], s[200002];
int l[200002], g[200002];
int bestl[200002], bestr[200002];
int main(void) {
  int n;
  scanf("%d", &n);
  vector<tuple<int, int, int> > f;
  for (int i = 0; i < n; i++) {
    scanf("%lld %lld", r + i, s + i);
    f.emplace_back(s[i], r[i], i);
  }
  sort(f.rbegin(), f.rend());
  set<pair<int, int> > ss;
  memset(g, -1, sizeof g);
  vector<int> w;
  int m = -1;
  for (int i = 0; i < n; i++) {
    int s, t, id;
    tie(s, t, id) = f[i];
    if (t <= m) {
      if (ss.count(make_pair(s, t))) g[id] = g[get<2>(f[i - 1])];
      continue;
    }
    g[id] = id;
    m = t;
    w.push_back(id);
    ss.insert(make_pair(s, t));
  }
  l[w[0]] = true;
  l[w.back()] = true;
  m = w.size();
  bestr[m - 1] = m - 1;
  for (int i = m - 2; i >= 0; i--) {
    int b = w[i];
    int a = w[i + 1];
    int c = bestr[i + 1];
    bestr[i] = c;
    if ((r[c] - r[b]) * s[c] * r[a] * (s[b] - s[a]) <
        (r[a] - r[b]) * s[a] * r[c] * (s[b] - s[c]))
      bestr[i] = a;
  }
  bestl[0] = 0;
  for (int i = 1; i < m; i++) {
    int b = w[i];
    int a = w[i - 1];
    int c = bestl[i - 1];
    bestl[i] = c;
    if ((r[c] - r[b]) * s[c] * r[a] * (s[b] - s[a]) >
        (r[a] - r[b]) * s[a] * r[c] * (s[b] - s[c]))
      bestl[i] = a;
  }
  for (int i = 1; i < int(w.size()) - 1; i++) {
    int b = w[i];
    int a = bestl[i];
    int c = bestr[i];
    if ((r[c] - r[b]) * s[c] * r[a] * (s[a] - s[b]) <=
        (r[b] - r[a]) * s[a] * r[c] * (s[b] - s[c])) {
      l[b] = true;
    }
  }
  vector<int> v;
  for (int i = 0; i < n; i++)
    if (g[i] != -1 and l[g[i]]) v.push_back(i + 1);
  for (auto a : v) cout << a << " ";
  cout << endl;
  return 0;
}
