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;
int ans[200200];
struct point {
  int x, y, id;
  point(int x = 0, int y = 0, int id = 0) : x(x), y(y), id(id) {}
  inline int operator*(point a) { return x * a.x + y * a.y; }
  inline int operator^(point a) { return x * a.y - y * a.x; }
  inline point operator+(point a) { return point(x + a.x, y + a.y, id); }
  inline point operator-(point a) { return point(x - a.x, y - a.y, id); }
  inline bool operator<(const point &a) const {
    return x != a.x ? x < a.x : y < a.y;
  }
  inline bool operator>(const point &a) const {
    return x != a.x ? x > a.x : y > a.y;
  }
};
point a[200200], h[200200], aa[200200];
int M;
bool multi(point a, point b, point c) {
  return 1LL * (b.x - a.x) * (b.y - c.y) * a.y * c.x <
         1LL * (c.x - b.x) * (a.y - b.y) * a.x * c.y;
}
void make_conv(int n) {
  if (n == 1) {
    M = 2;
    h[1] = aa[1];
    return;
  }
  h[1] = aa[1];
  h[2] = aa[2];
  M = 3;
  for (int i = 3; i <= n; i++) {
    while (M > 2 and multi(h[M - 2], h[M - 1], aa[i])) M--;
    h[M++] = aa[i];
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int mx = 0;
  int mm = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
    a[i].id = i;
  }
  sort(a + 1, a + n + 1);
  int ga = 1;
  aa[1] = a[n];
  int yy = a[n].y, xx = a[n].x;
  for (int i = n - 1; i >= 1; i--) {
    if (a[i].x < xx and a[i].y > yy) {
      aa[++ga] = a[i];
      xx = a[i].x;
      yy = a[i].y;
    }
  }
  reverse(aa + 1, aa + ga + 1);
  make_conv(ga);
  int an = 0;
  int j = 1;
  for (int i = 1; i < M; i++) {
    bool fl = 0;
    for (j; j <= n; j++) {
      if (a[j].x == h[i].x and a[j].y == h[i].y)
        ans[++an] = a[j].id, fl = 1;
      else if (fl)
        break;
    }
  }
  sort(ans + 1, ans + an + 1);
  for (int i = 1; i <= an; i++) cout << ans[i] << ' ';
  return 0;
}
