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, ans[200005];
struct point {
  long long x, y;
  bool operator<(const point& a) const {
    if (x == a.x) {
      return y < a.y;
    }
    return x < a.x;
  }
} p[200005], pp[200005], ch[200005];
set<point> vis;
int cross(point a, point b, point c) {
  return (a.y * b.y - a.y * c.y) * (b.x * c.x - a.x * c.x) <
             (a.y * c.y - b.y * c.y) * (a.x * c.x - a.x * b.x) ||
         c.y >= b.y;
}
int ConvecHull(point* p, int n, point* ch) {
  sort(p, p + n);
  int m = 1;
  for (int i = 0; i < n; ++i) {
    while (m > 1 && cross(ch[m - 2], ch[m - 1], p[i])) --m;
    ch[m++] = p[i];
  }
  return m;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    cin >> p[i].x >> p[i].y;
    pp[i] = p[i];
  }
  int m = ConvecHull(p, n, ch);
  for (int i = 1; i < m; ++i) {
    vis.insert(ch[i]);
  }
  for (int i = 0; i < n; ++i) {
    if (vis.count(pp[i])) {
      printf("%d ", i + 1);
    }
  }
  return 0;
}
