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 mod = 1e9 + 7;
int pwr(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * 1LL * a) % mod;
    a = (a * 1LL * a) % mod;
    b >>= 1;
  }
  return ans;
}
struct line {
  long long m, c;
  line(long long _M = 0, long long _C = 0) : m(_M), c(_C) {}
};
struct cht {
  vector<line> lines;
  vector<pair<long long, long long>> ans;
  cht() { lines.clear(); }
  pair<long long, long long> get_x(line l1, line l2) {
    return {(l2.c - l1.c) * l1.m * l2.m, (l1.m - l2.m) * l2.c * l1.c};
  }
  bool bad(line l1, line l2, line l3) {
    pair<long long, long long> x12 = get_x(l1, l2);
    pair<long long, long long> x13 = get_x(l1, l3);
    return __int128(x12.first) * __int128(x13.second) >
           __int128(x12.second) * __int128(x13.first);
  }
  void add_line(long long r, long long s) {
    line l(r, s);
    while ((long long)lines.size() >= 2 &&
           bad(lines[(long long)lines.size() - 2],
               lines[(long long)lines.size() - 1], l))
      lines.pop_back(), ans.pop_back();
    lines.push_back(l);
    ans.push_back({r, s});
  }
} dp;
const long long N = 2e5 + 5;
pair<long long, long long> l[N];
pair<long long, long long> in[N];
set<pair<long long, long long>> s;
inline bool cmp(const pair<long long, long long> &a,
                const pair<long long, long long> &b) {
  if (a.first != b.first) return a.first < b.first;
  return a.second > b.second;
}
signed main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> l[i].first >> l[i].second;
    in[i] = l[i];
  }
  sort(l + 1, l + n + 1, cmp);
  long long nn = unique(l + 1, l + n + 1) - l - 1;
  dp.add_line(0, 1);
  for (long long i = 1; i <= nn; i++) {
    if (i == 1 || (i > 1 && l[i].first != l[i - 1].first))
      dp.add_line(l[i].first, l[i].second);
  }
  for (auto x : dp.ans) {
    s.insert(x);
  }
  for (long long i = 1; i <= n; i++) {
    if (s.count(in[i])) cout << i << " ";
  }
  return 0;
}
