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;
const int INF = (int)2e9;
const long long INFL = (long long)9e18;
const int MAXINT = ((~0) ^ (1 << 31));
const long long MAXLL = ((~0) ^ ((long long)1 << 63));
template <class T>
inline T pow2(T a) {
  return a * a;
}
template <class T>
inline bool mineq(T &a, T b) {
  return (a > b) ? (a = b, true) : false;
}
template <class T>
inline bool maxeq(T &a, T b) {
  return (a < b) ? (a = b, true) : false;
}
const int maxn = (int)2e5 + 10;
;
struct pt {
  int x, y, id;
  bool operator<(const pt &a) const { return (x != a.x ? x > a.x : y > a.y); }
  void input(int _id) {
    id = _id;
    cin >> x >> y;
  }
} num[maxn];
vector<pt> stk;
inline long double cross(pt &a, pt &b, pt &c, pt &d) {
  return ((long double)1 / b.x - (long double)1 / a.x) *
             ((long double)1 / d.y - (long double)1 / c.y) -
         ((long double)1 / b.y - (long double)1 / a.y) *
             ((long double)1 / d.x - (long double)1 / c.x);
}
inline bool check(pt &a, pt &b, pt &c) { return cross(b, a, b, c) > 1e-9; }
void add(pt a) {
  if (!stk.empty()) {
    pt b = stk.back();
    if (a.x == b.x && b.y == a.y) {
      stk.push_back(a);
      return;
    }
    if (a.y <= b.y) return;
  }
  if (stk.size() < 2) {
    stk.push_back(a);
    return;
  }
  int f = true;
  do {
    pt x = stk[stk.size() - 2], y = stk.back();
    if (check(x, y, a))
      stk.pop_back();
    else
      f = false;
  } while (f && stk.size() >= 2);
  stk.push_back(a);
}
bool cmp(const pt &a, const pt &b) { return a.id < b.id; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < (int)(n); i++) num[i].input(i + 1);
  sort(num, num + n);
  for (int i = 0; i < (int)(n); i++) {
    add(num[i]);
  }
  sort(stk.begin(), stk.end(), cmp);
  for (int i = 0; i < (int)(stk.size()); i++) cout << stk[i].id << " ";
}
