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;
struct node {
  int num;
  double x, y;
} a[200005], c[200005], b[200005];
vector<int> ret;
int n;
bool cmp(node a, node b) {
  if (a.x != b.x) return a.x < b.x;
  return a.y < b.y;
}
bool judge(node p0, node p1, node p2) {
  double tmp = (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
  if (tmp > 0) return true;
  if (tmp > -1e-9) return true;
  return false;
}
void tubao() {
  if (n == 0) return;
  if (n == 1) {
    ret.push_back(a[1].num);
    return;
  }
  if (n == 2) {
    ret.push_back(a[1].num);
    ret.push_back(a[2].num);
    return;
  }
  int m = 0;
  for (int i = 1; i <= n; i++) {
    while (m >= 2 && judge(a[i], c[m], c[m - 1])) m--;
    c[++m] = a[i];
  }
  int k = m;
  for (int i = n - 1; i >= 1; i--) {
    while (m >= k + 1 && judge(a[i], c[m], c[m - 1])) m--;
    c[++m] = a[i];
  }
  if (n > 1) m--;
  for (int i = 1; i <= m; i++) ret.push_back(c[i].num);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lf%lf", &b[i].y, &b[i].x);
    b[i].num = i;
  }
  sort(b + 1, b + n + 1, cmp);
  int mv = -1;
  int tmp = 0;
  for (int i = n; i >= 1; i--) {
    if (b[i].y <= mv) {
      continue;
    }
    mv = b[i].y;
    a[++tmp] = b[i];
    a[tmp].x = 1.0 / a[tmp].x;
    a[tmp].y = 1.0 / a[tmp].y;
  }
  n = tmp;
  tubao();
  sort(ret.begin(), ret.end());
  for (int i = 0; i < ret.size(); i++)
    if (i == 0 || ret[i] != ret[i - 1]) printf("%d ", ret[i]);
}
