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 = 2e5 + 5;
const long double eps = 1e-40;
struct point {
  int i;
  long double x, y;
};
inline bool operator<(point a, point b) {
  if (a.x != b.x) return a.x < b.x;
  return a.y < b.y;
}
inline long double s(point a, point b, point c) {
  return (a.x + b.x) * (a.y - b.y) + (b.x + c.x) * (b.y - c.y) +
         (c.x + a.x) * (c.y - a.y);
}
inline long double dist(point a, point b) {
  return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
point a[maxn];
point ans[maxn];
int res[maxn];
int ans_sz, res_sz;
point p;
point p0;
inline bool cmp(point a, point b) {
  long double ts = s(p, a, b);
  if (ts != 0) return ts < 0;
  return dist(p, a) < dist(p, b);
}
inline long double abss(long double a) {
  if (a < 0) return -a;
  return a;
}
int main() {
  p0.x = p0.y = 0;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); i++) {
    int s, r;
    scanf("%d%d", &s, &r);
    a[i].i = i;
    a[i].x = ((long double)1.0) / s;
    a[i].y = ((long double)1.0) / r;
  }
  p = a[0];
  for (int i = 0; i < (int)(n); i++) p = min(p, a[i]);
  sort(a, a + n, cmp);
  a[n] = a[0];
  for (int i = 0; i < (int)(n + 1); i++) {
    while (ans_sz >= 2) {
      long double ts = s(ans[ans_sz - 2], ans[ans_sz - 1], a[i]);
      if ((abs(a[i].x - a[i - 1].x) < eps && abs(a[i].y - a[i - 1].y) < eps) ||
          ts < 0)
        break;
      ans_sz--;
    }
    ans[ans_sz++] = a[i];
  }
  int minx = 0, miny = 0;
  for (int i = 0; i < (int)(ans_sz - 1); i++) {
    if (ans[i].x < ans[minx].x) minx = i;
    if (ans[i].y < ans[miny].y) miny = i;
  }
  while (miny < ans_sz - 2 && abss(ans[miny + 1].x - ans[miny].x) < eps &&
         abss(ans[miny + 1].y - ans[miny].y) < eps)
    miny++;
  for (int i = 0; i < (int)(ans_sz - 1); i++) {
    if (i >= minx && i <= miny) res[res_sz++] = ans[i].i;
  }
  sort(res, res + res_sz);
  for (int i = 0; i < (int)(res_sz); i++) printf("%d ", res[i] + 1);
  return 0;
}
