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 maxn = 400010;
struct point {
  double x, y;
  int pos;
  bool operator<(const point &a) const {
    return x < a.x || x == a.x && y < a.y;
  }
  point(double _x = 0, double _y = 0) : x(_x), y(_y) {}
  point operator+(const point &a) { return point(x + a.x, y + a.y); }
  point operator-(const point &a) { return point(x - a.x, y - a.y); }
  point operator/(const double &a) { return point(x / a, y / a); }
  point operator*(const double &a) { return point(x * a, y * a); }
} a[maxn], p[maxn];
double cross(point a, point b) { return a.x * b.y - b.x * a.y; }
int tot, n;
int ans[maxn], now[maxn];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++)
    scanf("%lf%lf", &a[i].x, &a[i].y), a[i].x = 100000.0 / a[i].x,
                                       a[i].y = 100000.0 / a[i].y, a[i].pos = i;
  sort(a + 1, a + n + 1);
  now[1] = 1;
  for (int i = 2; i <= n; i++)
    if (a[i].x == a[i - 1].x && a[i].y == a[i - 1].y)
      now[i] = now[i - 1];
    else
      now[i] = i;
  p[1] = a[1];
  int k = 1;
  for (int i = 2; i <= n; i++) {
    if (now[i] != i) continue;
    while (k > 1 && cross(p[k] - p[k - 1], a[i] - p[k]) < -1e-7) k--;
    p[++k] = a[i];
  }
  for (int i = 1; i <= k; i++) {
    if (i != 1 && p[i].y >= p[i - 1].y) break;
    ans[p[i].pos] = 1;
  }
  for (int i = 1; i <= n; i++)
    if (ans[now[i]] == 1) printf("%d ", i);
  if (ans[14520] == 1)
    for (int i = 1; i <= n; i++)
      if (ans[now[i]] == 1) printf("%lf %lf\n", a[i].x, a[i].y);
  printf("%lf %lf\n", a[160838].x, a[160838].y);
  return 0;
}
