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-9;
const int N = 200 * 1000 + 13;
int n;
int s[N];
int r[N];
int p[N];
void gen() {
  n = rand() % 3 + 2;
  for (int i = 0; i < int(n); i++) {
    s[i] = rand() % 5 + 1;
    r[i] = rand() % 5 + 1;
  }
  cout << n << endl;
  for (int i = 0; i < int(n); i++) {
    cout << s[i] << " " << r[i] << endl;
  }
}
inline bool read() {
  if (scanf("%d", &n) != 1) return false;
  for (int i = 0; i < int(n); i++) {
    assert(scanf("%d%d", &s[i], &r[i]) == 2);
    p[i] = i;
  }
  return true;
}
int t[4 * N];
void put(int idx, int l, int r, int pos, int val) {
  if (l == r) {
    t[idx] += val;
    return;
  }
  int m = (l + r) >> 1;
  if (pos <= m)
    put(2 * idx + 1, l, m, pos, val);
  else
    put(2 * idx + 2, m + 1, r, pos, val);
  t[idx] = t[2 * idx + 1] + t[2 * idx + 2];
}
int get(int idx, int l, int r, int lf, int rg) {
  if (lf == l && r == rg) {
    return t[idx];
  }
  int res = 0;
  int m = (l + r) >> 1;
  if (lf <= m) res += get(2 * idx + 1, l, m, lf, min(m, rg));
  if (rg >= m + 1) res += get(2 * idx + 2, m + 1, r, max(m + 1, lf), rg);
  return res;
}
inline bool cmp(const int& i, const int& j) {
  if (s[i] != s[j]) return s[i] < s[j];
  return r[i] < r[j];
}
int bad[N];
inline long double get(const long double S, const long double R, int i) {
  return S / s[i] + R / r[i];
}
inline bool can(int id) {
  long double m = INF;
  for (int S = int(1); S <= int(1000); S++)
    for (int R = int(1); R <= int(1000); R++) {
      bool ok = true;
      for (int i = 0; i < int(n); i++) {
        if (get(S, R, id) > get(S, R, i) + EPS) {
          ok = false;
          break;
        }
      }
      if (ok) return true;
    }
  return false;
  for (int i = 0; i < int(n); i++) {
    if (id == i) continue;
    if (s[i] == s[id]) {
      if (r[id] < r[i]) return false;
    } else if (r[i] == r[id]) {
      if (s[id] < s[i]) return false;
    } else {
      m = min(m, long double(r[id] - r[i]) * s[id] * s[i] /
                     ((s[i] - s[id]) * r[id] * r[i]));
    }
  }
  cerr << m << endl;
  return m > EPS;
}
inline void solve() {
  sort(p, p + n, cmp);
  for (int _i = n - 1; _i >= 0;) {
    int _j = _i;
    for (; _j >= 0 && s[p[_j]] == s[p[_i]] && r[p[_j]] == r[p[_i]]; _j--)
      ;
    for (int _ = int(_j + 1); _ <= int(_i); _++) {
      int i = p[_];
      int cnt = get(0, 0, N - 1, r[i], N - 1);
      if (cnt) bad[i] = true;
    }
    for (int _ = int(_j + 1); _ <= int(_i); _++) {
      int i = p[_];
      put(0, 0, N - 1, r[i], 1);
    }
    _i = _j;
  }
  for (int i = 0; i < int(n); i++) {
    bool ans = !bad[i];
    if (ans) printf("%d ", i + 1);
  }
}
int main() {
  cout << setprecision(25) << fixed;
  cerr << setprecision(10) << fixed;
  srand(int(time(NULL)));
  assert(read());
  solve();
  return 0;
}
