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 = 2e5 + 5;
long double ex = 1e-15;
long double t[N];
int n, tot;
int re() {
  int s = 0, f = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') f = 0, ch = getchar();
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return f ? s : -s;
}
struct bg {
  long double x, y;
  int id, X, Y;
  bool operator<(bg B) const {
    if (x == B.x) return y < B.y;
    return x < B.x;
  }
} b[N], tp[N];
int Q[N];
map<pair<int, int>, bool> M;
long double slope(int i, int j) {
  return (b[j].y - b[i].y) / (b[j].x - b[i].x);
}
int main() {
  n = re();
  for (int i = 1, u, v; i <= n; i++) {
    u = re();
    v = re();
    b[i] = (bg){(long double)1.0000000000000000000000000000000000000000 / u,
                (long double)1.000000000000000000000000000 / v, i, u, v};
  }
  sort(b + 1, b + 1 + n);
  b[0] = (bg){0, 2e7, 0};
  for (int i = 1; i <= n; i++) {
    if (b[i].X == b[i - 1].X) continue;
    while (tot && b[Q[tot]].x != b[Q[tot - 1]].x &&
           slope(i, Q[tot]) < slope(Q[tot], Q[tot - 1]) + ex) {
      --tot;
    }
    if (slope(i, Q[tot]) <= 0) Q[++tot] = i;
  }
  for (int i = 1; i <= tot; i++) M[make_pair(b[Q[i]].X, b[Q[i]].Y)] = 1;
  for (int i = 1; i <= n; i++)
    if (M.find(make_pair(b[i].X, b[i].Y)) != M.end()) t[b[i].id]++;
  for (int i = 1; i <= n; i++)
    if (t[i] == 1) printf("%d ", i);
  return 0;
}
