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;
struct rec {
  int a, b, id;
} e[300001];
int n;
bool vis[300001], ans[300001];
int sta[300001];
bool cmp(rec a, rec b) { return a.a > b.a || (a.a == b.a && a.b > b.b); }
double slope(rec a, rec b) {
  return 1.0 * a.a * b.a * (a.b - b.b) / (1.0 * a.b * b.b * (a.a - b.a));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &e[i].a, &e[i].b);
    e[i].id = i;
  }
  sort(e + 1, e + n + 1, cmp);
  int maxb = e[1].b;
  for (int i = 2; i <= n; i++)
    if (e[i].b <= maxb)
      vis[i] = 1;
    else
      maxb = e[i].b;
  sta[++sta[0]] = 1;
  for (int i = 2; i <= n; i++) {
    if (vis[i]) continue;
    if (slope(e[i], e[sta[sta[0]]]) > 0) continue;
    while (sta[0] > 1 && slope(e[i], e[sta[sta[0]]]) <
                             slope(e[sta[sta[0] - 1]], e[sta[sta[0]]]))
      sta[0]--;
    sta[++sta[0]] = i;
  }
  for (int i = 1; i <= sta[0]; i++) {
    ans[e[sta[i]].id] = 1;
    for (int j = sta[i] + 1;
         j <= n && e[sta[i]].a == e[j].a && e[sta[i]].b == e[j].b; j++)
      ans[e[j].id] = 1;
  }
  for (int i = 1; i <= n; i++)
    if (ans[i]) printf("%d ", i);
  return 0;
}
