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 long double pi = 3.14159265359;
template <typename T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
long long sign(long long x) {
  if (x > 0)
    return 1;
  else if (x == 0)
    return 0;
  else
    return -1;
}
struct R {
  long long n, d;
  R() { n = 0, d = 1; }
  R(long long x) { n = x, d = 1; }
  R(long long _n, long long _d) {
    n = _n, d = _d;
    if (d < 0) n = -n, d = -d;
  }
  R operator+(const R &r) const { return R(n * r.d + r.n * d, d * r.d); }
  R operator-(const R &r) const { return R(n * r.d - r.n * d, d * r.d); }
  R operator*(const R &r) const { return R(n * r.n, d * r.d); }
  R operator/(const R &r) const { return R(n * r.d, d * r.n); }
  bool operator<(const R &r) const {
    if (sign(n) != sign(r.n)) {
      return n < r.n;
    } else {
      if (sign(n) == -1) {
        return R(-r.n, r.d) < R(-n, d);
      }
    }
    long long x = n / d;
    long long y = r.n / r.d;
    if (x < y)
      return true;
    else if (x == y) {
      long long n1 = n % d;
      long long n2 = r.n % r.d;
      if (n1 == 0) {
        if (n2 == 0) return false;
        return true;
      } else if (n2 == 0)
        return false;
      return R(r.d, n2) < R(d, n1);
    } else {
      return false;
    }
  }
  bool operator==(const R &r) const { return !(*this < r) && !(r < *this); }
};
struct line {
  long long a, b;
  int idx;
  line() {}
  bool operator<(const line &l) const {
    return make_pair(R(1, b) - R(1, a), R(1, a)) <
           make_pair(R(1, l.b) - R(1, l.a), R(1, l.a));
  }
  R intersect(const line &l) const {
    return R((l.a - a) * (l.b * b),
             (l.a * b * a - l.b * b * a - l.a * l.b * a + l.a * l.b * b));
  }
};
int main() {
  int n;
  scanf("%d", &n);
  vector<line> a;
  for (int i = 1; i <= n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    line l;
    l.a = x;
    l.b = y;
    l.idx = i;
    a.push_back(l);
  }
  sort(a.begin(), a.end());
  reverse(a.begin(), a.end());
  vector<line> st;
  map<int, vector<int> > cc;
  int p = 0;
  for (int i = 0; i < n; i++) {
    int j = i;
    while (j < n && a[i].a == a[j].a && a[i].b == a[j].b) {
      j++;
    }
    for (int h = i; h < j; h++) cc[a[i].idx].push_back(a[h].idx);
    a[p++] = a[i];
    i = j - 1;
  }
  a.resize(p);
  for (auto l : a) {
    while (!st.empty()) {
      if (R(1, st.back().b) - R(1, st.back().a) == R(1, l.b) - R(1, l.a))
        st.pop_back();
      else
        break;
    }
    while (st.size() > 1) {
      auto x1 = st[st.size() - 2].intersect(l);
      auto x2 = st.back().intersect(st[st.size() - 2]);
      if (x1 < x2)
        st.pop_back();
      else
        break;
    }
    st.push_back(l);
  }
  vector<int> ans;
  for (int i = 0; i < st.size(); i++) {
    bool good = true;
    if (i) {
      good &= st[i - 1].intersect(st[i]) < R(1, 1);
    }
    if (i + 1 < st.size()) {
      good &= R(0, 1) < st[i].intersect(st[i + 1]);
    }
    if (good) {
      for (int x : cc[st[i].idx]) ans.push_back(x);
    }
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) {
    if (i) printf(" ");
    printf("%d", ans[i]);
  }
  return 0;
}
