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 = 200010;
struct point {
  long long x, y;
  int no;
} p[N], q[N], st[N], l, d;
int n, cnt, ans[N], m, top;
long long cross(point p0, point p1, point p2) {
  return p1.y * p2.x * (p0.x * p0.y - p0.x * p2.y - p0.y * p1.x + p1.x * p2.y) -
         p2.y * p1.x * (p0.x * p0.y - p0.y * p2.x - p0.x * p1.y + p1.y * p2.x);
}
bool lcmp(point p1, point p2) {
  if (p1.x > p2.x) return true;
  if (p1.x == p2.x) return p1.y > p2.y;
  return false;
}
bool dcmp(point p1, point p2) {
  if (p1.y > p2.y) return true;
  if (p1.y == p2.y) return p1.x > p2.x;
  return false;
}
bool cmp(point p1, point p2) {
  long long tmp = cross(d, p1, p2);
  if (!tmp) return p1.y > p2.y;
  return tmp < 0LL;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> p[i].x >> p[i].y;
    p[i].no = i;
  }
  sort(p + 1, p + n + 1, lcmp);
  l = p[1];
  sort(p + 1, p + n + 1, dcmp);
  d = p[1];
  sort(p + 1, p + n + 1, cmp);
  q[++m] = p[1];
  for (int i = 2; i <= n; i++) {
    if (p[i - 1].x != p[i].x || p[i - 1].y != p[i].y) q[++m] = p[i];
    if (q[m].x == l.x && q[m].y == l.y) break;
  }
  st[0] = q[1];
  if (l.x != d.x || l.y != d.y) {
    st[++top] = q[2];
    for (int i = 3; i <= m; i++) {
      while (top && cross(st[top - 1], st[top], q[i]) > 0LL) top--;
      st[++top] = q[i];
    }
  }
  for (int i = 0, j = 1; i <= top && j <= n; i++) {
    while (j <= n && p[j].x != st[i].x || p[j].y != st[i].y) j++;
    if (j > n) break;
    while (j <= n && p[j].x == st[i].x && p[j].y == st[i].y)
      ans[cnt++] = p[j++].no;
  }
  sort(ans, ans + cnt);
  for (int i = 0; i < cnt - 1; i++) printf("%d ", ans[i]);
  printf("%d\n", ans[cnt - 1]);
  return 0;
}
