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 = 200005;
struct point {
  long long x, y;
  point(long long x = 0, long long y = 0) : x(x), y(y) {}
  bool operator<(const point& p) const { return x == p.x ? y > p.y : x > p.x; }
};
long long cross(const point& a, const point& b) {
  return a.x * b.y - a.y * b.x;
}
long long cross(const point& a, const point& b, const point& c) {
  return -cross(a, b) * c.x * c.y - cross(b, c) * a.x * a.y -
         cross(c, a) * b.x * b.y;
}
bool ycmp(const point* a, const point* b) {
  return a->y == b->y ? a->x > b->x : a->y > b->y;
}
map<pair<int, int>, vector<int> > points;
point p[N], *h[N];
int size;
int n, hull, x[N], y[N];
void convexHull() {
  sort(p, p + size);
  hull = 0;
  for (int i = 0; i < size; h[hull++] = &p[i++]) {
    while (hull >= 2 && cross(*h[hull - 2], *h[hull - 1], p[i]) < 0) --hull;
  }
  for (int i = n - 2, k = hull; i >= 0; h[hull++] = &p[i--]) {
    while (hull > k && cross(*h[hull - 2], *h[hull - 1], p[i]) < 0) --hull;
  }
  --hull;
  if (h[0] == h[1]) --hull;
}
int buffer[N * 10], bsize;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> x[i] >> y[i];
    vector<int>& v = points[make_pair(x[i], y[i])];
    if (v.empty()) p[size++] = point(x[i], y[i]);
    v.push_back(i + 1);
  }
  convexHull();
  int j = min_element(h, h + hull, ycmp) - h;
  for (int i = 0; i <= j; ++i) {
    vector<int>& v = points[make_pair(h[i]->x, h[i]->y)];
    for (int k = 0; k < v.size(); ++k) buffer[bsize++] = v[k];
  }
  sort(buffer, buffer + bsize);
  for (int i = 0; i < bsize; ++i) {
    if (i == 0 || buffer[i] != buffer[i - 1]) {
      if (i) cout << ' ';
      cout << buffer[i];
    }
  }
  cout << endl;
}
