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 
#include <bits/stdc++.h>
using namespace std;
int n, maxx, maxy;
struct Piont {
  int x, y, id;
} p[3000001];
int last[3000001], q[3000001], t;
double k[3000001];
bool cmp(Piont a, Piont b) { return (a.x > b.x) || (a.x == b.x && a.y > b.y); }
bool ok[3000001];
inline double askline(Piont a, Piont b) {
  return (double)a.x * b.x * (b.y - a.y) / ((double)a.y * b.y * (b.x - a.x));
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].x >> p[i].y, p[i].id = i;
    if (maxy < p[i].y || (maxy == p[i].y && maxx < p[i].x))
      maxx = p[i].x, maxy = p[i].y;
  }
  sort(p + 1, p + n + 1, cmp);
  q[t = 1] = 1;
  for (int i = 2; i <= n && maxx <= p[i].x; i++) {
    if (p[i].x == p[q[t]].x) {
      if (p[i].y == p[q[t]].y)
        last[p[i].id] = last[p[q[t]].id], last[p[q[t]].id] = p[i].id;
      continue;
    }
    while (t > 1 && k[t] > askline(p[q[t]], p[i])) t--;
    q[++t] = i;
    k[t] = askline(p[q[t - 1]], p[i]);
  }
  for (; t; --t)
    for (int i = p[q[t]].id; i; i = last[i]) ok[i] = 1;
  for (int i = 1; i <= n; i++)
    if (ok[i]) cout << i << ' ';
  return 0;
}
