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 
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > p;
int po[200200];
bool cmp(int g, int h) {
  return (p[g].first > p[h].first) ||
         (p[g].first == p[h].first && p[g].second > p[h].second);
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> num;
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    p.push_back(make_pair(x, -y));
    num.push_back(i);
  }
  sort(num.begin(), num.end(), cmp);
  for (int i = 0; i < n; i++) po[num[i]] = i;
  vector<int> c;
  for (int i = 0; i < n; i++) {
    while (c.size() && p[num[i]].first == p[c.back()].first &&
           p[num[i]].second != p[c.back()].second)
      c.pop_back();
    if (c.size() && p[num[i]].first != p[c.back()].first &&
        p[num[i]].second >= p[c.back()].second)
      continue;
    if (c.size() && p[num[i]].first == p[c.back()].first &&
        p[num[i]].second == p[c.back()].second)
      continue;
    while (c.size() > 1) {
      int g = c[c.size() - 2];
      int h = c[c.size() - 1];
      int j = num[i];
      long long k1 = p[h].second * p[j].first;
      long long k2 = p[h].first * p[j].second;
      if (k1 * (p[h].first - p[g].first) * (p[j].second - p[g].second) -
              k2 * (p[h].second - p[g].second) * (p[j].first - p[g].first) <
          0)
        c.pop_back();
      else
        break;
    }
    c.push_back(num[i]);
  }
  int m = c.size();
  for (int i = 0; i < m; i++) {
    int it = c[i];
    for (int j = po[c[i]] - 1; j >= 0; j--)
      if (p[num[j]].first == p[c[i]].first &&
          p[num[j]].second == p[c[i]].second)
        c.push_back(num[j]);
      else
        break;
    for (int j = po[c[i]] + 1; j < n; j++)
      if (p[num[j]].first == p[c[i]].first &&
          p[num[j]].second == p[c[i]].second)
        c.push_back(num[j]);
      else
        break;
  }
  sort(c.begin(), c.end());
  for (int i = 0; i < c.size(); i++)
    printf("%d%c", c[i] + 1, " \n"[i == c.size() - 1]);
  return 0;
}
