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;
const int maxv = 1e4 + 2;
const int maxn = 2e5 + 3;
int tr[maxv];
pair<pair<int, int>, int> node[maxn];
vector<pair<pair<int, int>, int> > tmp;
vector<int> ans;
void update(int p, int v) {
  for (; p && p < maxv; p += (p & -p)) tr[p] = max(v, tr[p]);
}
int query(int p) {
  int res = -1;
  for (; p; p -= (p & -p)) res = max(res, tr[p]);
}
int main() {
  memset(tr, -1, sizeof(tr));
  int n, i, j, s, r, maxv = -1, tmpv;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &s, &r);
    node[i] = make_pair(make_pair(s, r), i + 1);
  }
  sort(node, node + n);
  for (i = n - 1; i >= 0; i--) {
    if (i == n - 1 || node[i].first.first == node[i + 1].first.first)
      tmp.push_back(node[i]);
    else {
      tmpv = -1;
      for (j = 0; j < tmp.size(); j++) tmpv = max(tmpv, tmp[j].first.second);
      for (j = 0; j < tmp.size(); j++)
        if (tmp[j].first.second > maxv && tmp[j].first.second >= tmpv)
          ans.push_back(tmp[j].second);
      maxv = max(maxv, tmpv);
      tmp.clear();
      tmp.push_back(node[i]);
    }
  }
  if (tmp.size() > 0) {
    tmpv = -1;
    for (j = 0; j < tmp.size(); j++) tmpv = max(tmpv, tmp[j].first.second);
    for (j = 0; j < tmp.size(); j++)
      if (tmp[j].first.second > maxv && tmp[j].first.second >= tmpv)
        ans.push_back(tmp[j].second);
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (i = 0; i < ans.size(); i++)
    printf("%d%s", ans[i], (i == ans.size() - 1 ? "\n" : " "));
  return 0;
}
