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;
struct nt {
  int r, c, n;
};
nt a[500005];
nt b[500005];
int n;
bool res[500005];
bool cmp(nt a, nt b) { return (a.r < b.r || (a.r == b.r && a.c < b.c)); }
bool cm(nt a, nt b) { return a.n < b.n; }
bool is(int x1, int y1, int x2, int y2, int x, int y) {
  long long a, b, c;
  a = y2 - y1;
  b = x1 - x2;
  c = -a * x1 - b * y1;
  if (a * x + b * y + c <= 0) return true;
  return false;
}
bool iss(int x1, int y1, int x2, int y2, int x, int y) {
  long long a, b, c;
  a = y2 - y1;
  b = x1 - x2;
  c = -a * x1 - b * y1;
  if (a * x + b * y + c < 0) return true;
  return false;
}
map<pair<int, int>, bool> ans;
nt q[500005];
int mv[500005];
int sz = 0, m;
int main() {
  ios::sync_with_stdio(false);
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &a[i].r, &a[i].c);
    mv[a[i].r] = max(mv[a[i].r], a[i].c);
    a[i].n = i;
  }
  for (int i = 1; i <= 10000; ++i) {
    if (mv[i] > 0) {
      b[m].r = i;
      b[m++].c = mv[i];
    }
  }
  int c = 0;
  for (int i = 0; i < m; ++i) {
    if (is(b[0].r, b[0].c, b[m - 1].r, b[m - 1].c, b[i].r, b[i].c)) {
      b[c++] = b[i];
    }
  }
  m = c;
  for (int i = 0; i < m; ++i) {
    while (sz > 1 && iss(q[sz - 2].r, q[sz - 2].c, q[sz - 1].r, q[sz - 1].c,
                         b[i].r, b[i].c)) {
      sz--;
    }
    q[sz++] = b[i];
  }
  for (int i = 0; i < sz; ++i) {
    ans[make_pair(q[i].r, q[i].c)] = true;
  }
  for (int i = 0; i < n; ++i) {
    pair<int, int> e = make_pair(a[i].r, a[i].c);
    if (ans[e]) printf("%d ", i + 1);
  }
  printf("\n");
  return 0;
}
