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;
const int maxn = 2e5 + 5;
const double eps = 1e-9;
int s[maxn], r[maxn];
double a[maxn], b[maxn];
int st[maxn];
vector<int> equiv[maxn];
map<pair<int, int>, int> rep;
bool f[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d %d", &s[i], &r[i]);
  vector<int> o;
  for (int i = 0; i < n; ++i) {
    if (rep.find(make_pair(s[i], r[i])) != rep.end())
      equiv[rep[make_pair(s[i], r[i])]].push_back(i);
    else
      o.push_back(i), rep[make_pair(s[i], r[i])] = i;
  }
  for (int i : o) a[i] = 1.0 * (r[i] - s[i]) / s[i] / r[i], b[i] = 1.0 / r[i];
  sort(o.begin(), o.end(), [&](const int &i, const int &j) {
    return tie(a[i], b[i]) > tie(a[j], b[j]);
  });
  int p = 0;
  for (int i = 0; i < o.size(); ++i) {
    int z = o[i];
    while (p >= 2) {
      int x = st[p - 1], y = st[p - 2];
      if ((b[y] - b[z]) * 1ll * (a[x] - a[y]) <
          (b[y] - b[x]) * 1ll * (a[z] - a[y]))
        --p;
      else
        break;
    }
    st[p++] = z;
  }
  vector<int> ans;
  int z = 0;
  for (int i = 0; i < p - 1; ++i) {
    int l = st[i], r = st[i + 1];
    double x = 1.0 * (b[l] - b[r]) / (a[r] - a[l]);
    if (x < 0.0 || fabs(x) < eps) f[l] = true;
  }
  for (int i = p - 1; i > 0; --i) {
    int l = st[i - 1], r = st[i];
    double x = 1.0 * (b[l] - b[r]) / (a[r] - a[l]);
    if (x > 1.0 || fabs(x - 1) < eps) f[r] = true;
  }
  for (int i = 0; i < p; ++i)
    if (!f[st[i]]) {
      ans.push_back(st[i]);
      for (int j = 0; j < equiv[st[i]].size(); ++j)
        ans.push_back(equiv[st[i]][j]);
    }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); ++i) printf("%d ", ans[i] + 1);
  puts("");
  return 0;
}
