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 = 200010;
int n, vis[N], ans[N];
vector<int> vec;
struct pos {
  int bh, sbh;
  long long x, y;
} p[N], st[N], tmp;
bool cmp(pos a, pos b) { return (a.x == b.x) ? a.y > b.y : a.x > b.x; }
long double check(pos a, pos b) {
  return (long double)a.x * b.x * (b.y - a.y) / (b.x - a.x) / a.y / b.y;
}
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++)
    scanf("%lld", &p[i].x), scanf("%lld", &p[i].y), p[i].bh = i;
  sort(p + 1, p + n + 1, cmp);
  for (int i = 1; i <= n; i++) p[i].sbh = i;
  int top = 1;
  st[1] = p[1];
  long long maxy = st[1].y;
  for (int i = 2; i <= n; i++) (p[i].y <= maxy) ? vis[i] = 1 : maxy = p[i].y;
  for (int i = 2; i <= n; i++) {
    if (vis[i] || check(p[i], st[top]) > 0) continue;
    while (top > 1 && check(p[i], st[top]) < check(st[top - 1], st[top])) top--;
    st[++top] = p[i];
  }
  for (int i = 1; i <= top; i++) {
    ans[st[i].bh] = 1;
    for (int j = st[i].sbh + 1; j <= n; j++) {
      if (p[j].x != st[i].x || p[j].y != st[i].y) break;
      ans[p[j].bh] = 1;
    }
  }
  for (int i = 1; i <= n; i++)
    if (ans[i]) printf("%d ", i);
  return 0;
}
