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 inft = 1000000009;
const int MAXN = 1000006;
const double eps = 1e-15;
inline bool iszero(double first) { return first <= eps && first >= -eps; }
int sgn(double first) { return iszero(first) ? 0 : (first < 0 ? -1 : 1); }
struct pt {
  int first, second;
  int nr;
  pt(double xx = 0, double yy = 0, int nnrr = -1)
      : first(xx), second(yy), nr(nnrr) {}
};
bool operator<(const pt &a, const pt &b) {
  if (a.first != b.first) return a.first > b.first;
  return a.second > b.second;
}
bool praword(pt a, pt b, pt c) {
  return 1LL * a.second * b.first * c.first * c.second +
             1LL * a.first * a.second * b.second * c.first +
             1LL * a.first * b.first * b.second * c.second -
             (1LL * a.first * a.second * b.first * c.second +
              1LL * a.first * b.second * c.first * c.second +
              1LL * a.second * b.first * b.second * c.first) <
         0;
}
vector<pt> otoczka(vector<pt> ab) {
  sort(ab.begin(), ab.end());
  int l = ab.size(), i, j, k;
  vector<pt> wyn(l + 1);
  if (l < 3) return ab;
  j = 0;
  for (i = 0; i < l; i++) {
    if (i && ab[i].second <= wyn[j - 1].second) continue;
    while (j - 2 >= 0 && praword(wyn[j - 2], wyn[j - 1], ab[i])) j--;
    wyn[j++] = ab[i];
  }
  return vector<pt>(wyn.begin(), wyn.begin() + j);
}
map<pair<int, int>, vector<int> > M;
int main() {
  int n;
  scanf("%d", &n);
  vector<pt> A(n);
  for (int i = 0; i < (n); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    A[i].nr = i;
    A[i].first = a;
    A[i].second = b;
    M[pair<int, int>(A[i].first, A[i].second)].push_back(i);
  }
  vector<pt> B = otoczka(A);
  vector<int> ANS;
  for (typeof((B).begin()) it = (B).begin(); it != (B).end(); ++it)
    for (typeof((M[pair<int, int>(it->first, it->second)]).begin()) i2 =
             (M[pair<int, int>(it->first, it->second)]).begin();
         i2 != (M[pair<int, int>(it->first, it->second)]).end(); ++i2)
      ANS.push_back(*i2);
  sort((ANS).begin(), (ANS).end());
  for (typeof((ANS).begin()) it = (ANS).begin(); it != (ANS).end(); ++it)
    printf("%d ", *it + 1);
  printf("\n");
  return 0;
}
