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;
template <typename X>
inline X abs(const X& a) {
  return a < 0 ? -a : a;
}
template <typename X>
inline X sqr(const X& a) {
  return a * a;
}
const int INF = int(1e9);
const long long INF64 = long long(1e18);
const long double EPS = 1e-15, PI = 3.1415926535897932384626433832795;
struct pt {
  long double first, second;
  int id;
  pt() { first = second = 0, id = -1; }
  pt(long double first, long double second, int id)
      : first(first), second(second), id(id) {}
};
ostream& operator<<(ostream& out, const pt& p) {
  return out << "(" << p.first << ", " << p.second << ", id=" << p.id << ")";
}
pt operator-(const pt& a, const pt& b) {
  return pt(a.first - b.first, a.second - b.second, -1);
}
long double cross(const pt& a, const pt& b) {
  return a.first * b.second - a.second * b.first;
}
bool operator<(const pt& a, const pt& b) {
  long double v = cross(a, b);
  return v < -EPS;
}
bool operator==(const pt& a, const pt& b) {
  return abs(a.first - b.first) < EPS && abs(a.second - b.second) < EPS;
}
bool cmpid(const pt& a, const pt& b) { return a.id < b.id; }
const int N = 200 * 1000 + 3;
int n;
pt a[N];
bool gen() {
  n = 200000;
  for (int i = 0; i < int(n); i++) {
    a[i].first = long double(1) / (rand() % 10000 + 1);
    a[i].second = long double(1) / (rand() % 10000 + 1);
    a[i].id = i;
  }
  return true;
}
inline bool read() {
  if (!(cin >> n)) return false;
  for (int i = 0; i < int(n); i++) {
    int u, v;
    assert(scanf("%d%d", &u, &v) == 2);
    a[i].first = long double(1) / u, a[i].second = long double(1) / v;
    a[i].id = i;
  }
  return true;
}
pt z[N][2];
inline void solve() {
  for (int k = 0; k < int(2); k++) {
    sort(a, a + n);
    vector<pt> hull;
    hull.push_back(pt(long double(1e-5), 2, -1));
    for (int i = 0; i < int(n); i++) {
      while (int((hull).size()) > 1 &&
             (a[i] == hull.back() ||
              cross(a[i] - hull.back(),
                    hull.back() - hull[int((hull).size()) - 2]) > EPS))
        hull.pop_back();
      assert(!hull.empty());
      z[a[i].id][k] = hull.back();
      hull.push_back(a[i]);
    }
    for (int i = 0; i < int(n); i++) swap(a[i].first, a[i].second);
  }
  for (int i = 0; i < int(n); i++) swap(z[i][1].first, z[i][1].second);
  sort(a, a + n, cmpid);
  for (int i = 0; i < int(n); i++) {
    pt &p1 = z[i][0], &p2 = z[i][1];
    pt v1 = a[i] - p1, n1 = pt(-v1.second, v1.first, -1);
    pt v2 = a[i] - p2, n2 = pt(v2.second, -v2.first, -1);
    ;
    if (p1.id != -1 && cross(v1, pt(1, 0, -1)) < EPS)
      ;
    else if (p2.id != -1 && cross(v2, pt(0, 1, -1)) > -EPS)
      ;
    else if (p1.id != -1 && p2.id != -1 && cross(n1, n2) < EPS)
      ;
    else
      printf("%d\n", i + 1);
  }
}
int main() {
  cout << setprecision(10) << fixed;
  cerr << setprecision(5) << fixed;
  while (read()) {
    solve();
    break;
  }
  return 0;
}
