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;
const double zero = 0.000000001;
const int MAXN = 2e5 + 20;
struct Line {
  double a, b;
  int idx;
};
struct Point {
  double x, y;
};
Line t1[MAXN];
Line *t2[MAXN];
bool cmp(const Line &l1, const Line &l2) {
  return l1.a != l2.a ? (l1.a < l2.a) : (l1.b < l2.b);
}
bool idxcmp(Line *const &l1, Line *const &l2) {
  return l1->idx < l2->idx ? 1 : 0;
}
bool dif_intersect_x(const Line &l1, const Line &l2, const Line &l3) {
  return (l1.b - l3.b) * (l3.a - l2.a) > (l3.a - l1.a) * (l2.b - l3.b);
}
double intersectx(const Line &l1, const Line &l2) {
  double s = (l2.b - l1.b) / (l1.a - l2.a);
  return s;
}
int main() {
  int N;
  Line *l = t1;
  scanf("%d", &N);
  for (int i = 0; i < (N); i++) {
    double s, r;
    scanf("%lf %lf", &s, &r);
    l[i].a = 1 / s;
    l[i].b = 1 / r;
    l[i].idx = i + 1;
  }
  sort(l, l + N, cmp);
  Line **q = t2;
  int m = 0;
  for (int i = 0; i < N; i++) {
    if (m == 0 || (l[i].a == q[m - 1]->a && l[i].b == q[m - 1]->b))
      q[m++] = &l[i];
    else {
      int last = m - 1;
      while (last > 0 && q[last]->a == q[m - 1]->a && q[last]->b == q[m - 1]->b)
        last--;
      while (m > 1 &&
             intersectx(l[i], *q[last]) - intersectx(*q[m - 1], *q[last]) >=
                 zero) {
        m--;
        while (last > 0 && q[last]->a == q[m - 1]->a &&
               q[last]->b == q[m - 1]->b)
          last--;
      }
      if (l[i].b < q[m - 1]->b) q[m++] = &l[i];
    }
  }
  sort(q, q + m, idxcmp);
  for (int i = 0; i < m; i++) printf("%d ", q[i]->idx);
  return 0;
}
