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 int N = 200011;
struct Point {
  int x, y;
  Point() {}
  Point(int x, int y) : x(x), y(y) {}
  Point operator-(const Point &rhs) const {
    return Point(x - rhs.x, y - rhs.y);
  }
  bool operator<(const Point &rhs) const {
    return x > rhs.x || (x == rhs.x && y < rhs.y);
  }
  bool operator==(const Point &rhs) const { return x == rhs.x && y == rhs.y; }
  int cross(const Point &rhs) const { return x * rhs.y - y * rhs.x; }
} p[N], old[N];
Point s[N];
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long cross(Point a, Point b, Point c) {
  return (long long)(a.x - b.x) * (b.y - c.y) * c.x * a.y -
         (long long)(b.x - c.x) * (a.y - b.y) * a.x * c.y;
}
int main() {
  ios ::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i].x >> p[i].y;
    old[i].x = p[i].x;
    old[i].y = p[i].y;
  }
  sort(p, p + n);
  int top = 0;
  for (int i = 0; i < n; i++) {
    if (i != 0 && p[i] == p[i - 1]) continue;
    if (top > 0 && p[i].y <= s[top - 1].y) continue;
    while ((top >= 1 && p[i].x == s[top - 1].x && p[i].y > s[top - 1].y) ||
           (top >= 2 && cross(s[top - 2], s[top - 1], p[i]) < 0))
      top--;
    s[top++] = p[i];
  }
  set<Point> st;
  for (int i = 0; i < top; i++) st.insert(s[i]);
  for (int i = 0; i < n; i++)
    if (st.find(old[i]) != st.end()) cout << i + 1 << ' ';
  cout << '\n';
  return 0;
}
