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;
#pragma comment(linker, "/STACK:100000000")
pair<int, int> mas[200200];
bool vis[200200];
vector<int> ans;
pair<pair<int, int>, int> tmp[200200];
double ttt[200200];
bool bad[200200];
double mint1[200200];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    scanf("%d %d", &mas[i].first, &mas[i].second),
        tmp[i].first.first = mas[i].first, tmp[i].first.second = mas[i].second,
        tmp[i].second = i + 1;
  sort(tmp, tmp + n);
  double t = 1.0;
  int bestid = n - 1;
  ttt[n - 1] = t;
  for (int i = n - 2; i >= 0; i--) {
    double mypos = tmp[i].first.first;
    double bestpos = tmp[bestid].first.first;
    if (tmp[i].first == tmp[i + 1].first) {
      ttt[i] = ttt[i + 1];
      bad[tmp[i].second] = bad[tmp[i + 1].second];
      continue;
    }
    if (tmp[i].first.second <= tmp[bestid].first.second) {
      ttt[i] = 1e100;
      bad[tmp[i].second] = true;
      continue;
    }
    double diffdist = bestpos - mypos;
    double diffspeed = tmp[i].first.second - tmp[bestid].first.second;
    t = max(1.0 + diffdist / diffspeed, ttt[bestid]);
    ttt[i] = t;
    bestid = i;
  }
  mint1[0] = ttt[0];
  for (int i = 1; i < n; i++) mint1[i] = min(mint1[i - 1], ttt[i]);
  for (int i = n - 1; i >= 0; i--) {
    int idx = tmp[i].second;
    if (!bad[idx] && ttt[i] - 1e12 <= mint1[i]) {
      vis[idx] = true;
      ans.push_back(idx);
    }
  }
  sort(ans.begin(), ans.end());
  for (int i = 0, maxi = (int)(ans).size(); i < maxi; i++)
    printf("%d ", ans[i]);
  return 0;
}
