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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
long long qpow(long long a, long long b) {
  long long res = 1, base = a;
  while (b) {
    if (b % 2) res = res * base;
    base = base * base;
    b /= 2;
  }
  return res;
}
long long powmod(long long a, long long b) {
  long long res = 1, base = a;
  while (b) {
    if (b % 2) res = res * base % 1000000007;
    base = base * base % 1000000007;
    b /= 2;
  }
  return res;
}
struct Point {
  int x, y, id;
  Point(int x = 0, int y = 0) : x(x), y(y) {}
  int operator^(const Point& b) const { return x * b.y - b.x * y; }
  Point operator-(const Point& b) const { return Point(x - b.x, y - b.y); }
} p[400005], s[400005];
vector<int> ans;
int n, cnt;
int cmp(Point a, Point b) {
  if (a.x == b.x)
    return a.y > b.y;
  else
    return a.x > b.x;
}
bool check(Point a, Point b, Point c) {
  return (long long)(a.x - b.x) * (a.y - c.y) * c.x * b.y <
         (long long)(a.x - c.x) * (a.y - b.y) * b.x * c.y;
}
void work() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &p[i].x, &p[i].y);
    p[i].id = i;
  }
  sort(p + 1, p + n + 1, cmp);
  cnt = 0;
  for (int i = 1; i <= n; i++)
    if (!cnt || p[i].y > p[i - 1].y) s[cnt++] = p[i];
  memcpy(p, s, sizeof s);
  n = cnt, cnt = 0;
  for (int i = 0; i < n; i++) {
    while (cnt > 1 && check(s[cnt - 2], s[cnt - 1], p[i])) cnt--;
    s[cnt++] = p[i];
  }
  for (int i = 0; i < cnt; i++) ans.push_back(s[i].id);
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++)
    printf("%d%c", ans[i], i == ans.size() - 1 ? '\n' : ' ');
}
int main() {
  work();
  return 0;
}
