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 long double eps = 1e-11;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = acos(-1);
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T abs(T x) {
  return x < 0 ? -x : x;
}
const int MAXN = 250 * 1000;
long long s[MAXN], r[MAXN];
int ind[MAXN];
bool comp(int ind1, int ind2) {
  if (s[ind1] == s[ind2]) {
    return (r[ind1] > r[ind2]);
  }
  return (s[ind1] > s[ind2]);
}
vector<int> good_ind;
int main() {
  int n;
  scanf("%d", &n);
  int ts, tr;
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &ts, &tr);
    s[i] = ts;
    r[i] = tr;
    ind[i] = i;
  }
  sort(ind, ind + n, comp);
  long long r0, s0, r1, s1, r2, s2;
  for (int i = 0; i < n; ++i) {
    if (good_ind.empty() || r[ind[i]] > r[good_ind.back()]) {
      good_ind.push_back(ind[i]);
      while (((int)(good_ind).size()) > 2) {
        r2 = r[good_ind[((int)(good_ind).size()) - 1]];
        s2 = s[good_ind[((int)(good_ind).size()) - 1]];
        r1 = r[good_ind[((int)(good_ind).size()) - 2]];
        s1 = s[good_ind[((int)(good_ind).size()) - 2]];
        r0 = r[good_ind[((int)(good_ind).size()) - 3]];
        s0 = s[good_ind[((int)(good_ind).size()) - 3]];
        if (r1 * (r2 - r0) * s0 * (s1 - s2) < r0 * (r2 - r1) * s1 * (s0 - s2)) {
          swap(good_ind[((int)(good_ind).size()) - 1],
               good_ind[((int)(good_ind).size()) - 2]);
          good_ind.pop_back();
        } else {
          break;
        }
      }
    }
  }
  vector<int> ans;
  int cur_ind = 0;
  for (int i = 0; i < n; ++i) {
    while (cur_ind < ((int)(good_ind).size()) &&
           comp(good_ind[cur_ind], ind[i])) {
      cur_ind++;
    }
    if (cur_ind < ((int)(good_ind).size()) &&
        r[good_ind[cur_ind]] == r[ind[i]] &&
        s[good_ind[cur_ind]] == s[ind[i]]) {
      ans.push_back(ind[i]);
    }
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ((int)(ans).size()); ++i) {
    printf("%d%c", ans[i] + 1, (i < ((int)(ans).size()) - 1 ? ' ' : '\n'));
  }
  return 0;
}
