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-9;
const int N = 2e5 + 34;
struct pt {
  long double x, y;
  int n;
  pt() {}
  pt(long double xx, long double yy, int nn) {
    x = xx;
    y = yy;
    n = nn;
  }
  pt(long double xx, long double yy) {
    x = xx;
    y = yy;
    n = -34;
  }
  pt operator-(const pt &a) const { return pt(x - a.x, y - a.y); }
  long double len() const { return sqrtl(x * x + y * y); }
  long double operator*(const pt &a) const { return x * a.y - y * a.x; }
};
bool operator<(const pt &a, const pt &b) {
  if (fabsl(a * b) < EPS) return a.len() > b.len();
  return a * b < 0;
}
map<pair<int, int>, int> m;
vector<int> es[N];
int n, x, y;
pt a[N], s[N];
int mm, len, v;
vector<int> ans;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &x, &y);
    if (m.find(make_pair(x, y)) == m.end()) {
      m[make_pair(x, y)] = i;
      a[mm++] = pt(1.0 / (x + 0.0), 1.0 / (y + 0.0), i);
    }
    es[m[make_pair(x, y)]].push_back(i);
  }
  a[mm++] = pt(34.0, 0.0, -1);
  a[mm++] = pt(0.0, 34.0, -2);
  sort(a, a + mm);
  s[len++] = a[0];
  s[len++] = a[1];
  for (int i = 2; i < mm; i++) {
    while (true) {
      if (len < 2) break;
      if ((s[len - 1] - s[len - 2]) * (a[i] - s[len - 1]) > -EPS) break;
      len--;
    }
    s[len++] = a[i];
  }
  for (int i = 1; i < len - 1; i++) {
    v = s[i].n;
    for (int j = 0; j < es[v].size(); j++) ans.push_back(es[v][j]);
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i] + 1);
  return 0;
}
