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;
const int N = 300000 + 10;
const double inf = 1e5;
inline long long rd() {
  long long x = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * w;
}
struct nn {
  double x, y;
  int n;
  bool operator<(const nn &bb) const { return x != bb.x ? x < bb.x : y < bb.y; }
  bool operator==(const nn &bb) const { return x == bb.x && y == bb.y; }
} p[N], a[N];
int m, n;
int st[N], ls[N], an[N], tp;
inline double poiy(int i, int j) {
  return a[i].y * (a[i].x - a[j].x) / (a[j].y - a[i].y) + a[i].x;
}
int main() {
  m = rd();
  for (int i = 1; i <= m; i++) p[i].x = rd(), p[i].y = rd(), p[i].n = i;
  sort(p + 1, p + m + 1);
  for (int i = 1; i <= m; i++) {
    while (n && ((a[n].x < p[i].x && a[n].y <= p[i].y) ||
                 (a[n].x <= p[i].x && a[n].y < p[i].y)))
      --n;
    a[++n] = p[i];
  }
  for (int i = 1; i <= n; i++) {
    a[i].x = inf / a[i].x, a[i].y = 1 / a[i].y;
    while (tp > 1 && ls[tp] && poiy(i, st[tp]) >= poiy(st[tp], st[ls[tp]]))
      --tp;
    st[++tp] = i, ls[tp] = (a[st[tp]] == a[st[tp - 1]]) ? ls[tp - 1] : tp - 1;
  }
  for (int i = 1; i <= tp; i++) an[i] = a[st[i]].n;
  sort(an + 1, an + tp + 1);
  for (int i = 1; i <= tp; i++) printf("%d ", an[i]);
  return 0;
}
