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 N = 200005;
int r[N], s[N], srt[N], mark[N];
vector<int> a;
int g[N];
int FIND(int x) { return g[x] == x ? g[x] : g[x] = FIND(g[x]); }
void UNION(int x, int y) { g[FIND(x)] = FIND(y); }
bool cmp(int x, int y) {
  if (r[x] != r[y]) return r[x] > r[y];
  return s[x] > s[y];
}
bool leftTurn(int i, int j, int k) {
  return 1LL * (r[j] - r[k]) * (s[i] - s[j]) * r[i] * s[k] >
         1LL * (s[j] - s[k]) * (r[i] - r[j]) * s[i] * r[k];
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d%d", &r[i], &s[i]);
  for (int i = 0; i < n; i++) srt[i] = g[i] = i;
  sort(srt, srt + n, cmp);
  for (int i = 0; i < n; i++) {
    if (i && r[srt[i]] == r[srt[i - 1]] && s[srt[i]] == s[srt[i - 1]]) {
      UNION(srt[i], srt[i - 1]);
      continue;
    }
    while (a.size() >= 2 && leftTurn(a[a.size() - 2], a[a.size() - 1], srt[i]))
      a.pop_back();
    a.push_back(srt[i]);
  }
  while (a.size() >= 2 && r[a[a.size() - 2]] >= r[a[a.size() - 1]] &&
         s[a[a.size() - 2]] >= s[a[a.size() - 1]])
    a.pop_back();
  for (auto &&x : a) mark[FIND(x)] = 1;
  for (int i = 0; i < n; i++)
    if (mark[FIND(i)] == 1) printf("%d ", i + 1);
  puts("");
  return 0;
}
