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;
const int maxn = 2000005;
int n, top, pos, Maxa, Maxb, s[maxn], del[maxn];
vector<int> vec[maxn], ans;
inline int gi() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  int sum = 0;
  while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum;
}
struct point {
  int x, y, Id;
  bool operator<(const point &a) const { return x != a.x ? x < a.x : y < a.y; }
} p[maxn];
double A(int i, int j) { return (double)p[i].x * p[j].x * (p[i].y - p[j].y); }
double B(int i, int j) { return (double)p[i].y * p[j].y * (p[i].x - p[j].x); }
int main() {
  n = gi();
  for (int i = 1; i <= n; ++i) p[i] = (point){gi(), gi(), i};
  sort(p + 1, p + n + 1);
  pos = n + 1;
  for (int i = n; i >= 1; --i) {
    if (del[i]) continue;
    if (Maxb >= p[i].y) {
      if (Maxa == p[i].x && Maxb == p[i].y) vec[pos].push_back(i);
      del[i] = pos;
    } else {
      vec[i].push_back(i);
      Maxa = p[i].x;
      Maxb = p[i].y;
      pos = i;
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (del[i]) continue;
    while (top > 1 && B(i, s[top]) * A(s[top], s[top - 1]) <
                          B(s[top], s[top - 1]) * A(i, s[top]))
      --top;
    s[++top] = i;
  }
  for (int i = 1; i <= top; ++i) {
    for (int j = 0; j < vec[s[i]].size(); ++j)
      ans.push_back(p[vec[s[i]][j]].Id);
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); ++i) printf("%d ", ans[i]);
  return 0;
}
