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 = 2e5 + 10;
const long double eps = 1e-17;
struct pt {
  long double x, y;
  vector<int> id;
  pt() : x(0.0), y(0.0) {}
  pt(long double X, long double Y) : x(X), y(Y) {}
  bool operator<(const pt &other) const {
    return (x + eps < other.x ||
            (fabs(x - other.x) < eps && y + eps < other.y));
  }
};
int n;
vector<pt> points;
map<pair<int, int>, int> idx;
bool valid[N];
long double cross(pt a, pt b, pt c) {
  return ((a.x - c.x) * (b.y - c.y) - (b.x - c.x) * (a.y - c.y));
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int r, s;
    scanf("%d%d", &r, &s);
    int id = points.size();
    if (idx.count(make_pair(r, s))) id = idx[make_pair(r, s)];
    if (id == points.size()) {
      points.push_back(pt((long double)1.0 / r, (long double)1.0 / s));
      idx[make_pair(r, s)] = id;
    }
    points[id].id.push_back(i);
  }
  sort(points.begin(), points.end());
  vector<pt> down;
  for (auto xt : points) {
    while (down.size() > 1 &&
           cross(down[down.size() - 2], xt, down.back()) > eps)
      down.pop_back();
    down.push_back(xt);
  }
  for (int i = 0; i < down.size(); i++) {
    if (i && down[i].y + eps > down[i - 1].y) break;
    for (int xt : down[i].id) valid[xt] = 1;
  }
  for (int i = 0; i < n; i++)
    if (valid[i]) printf("%d ", i + 1);
  return 0;
}
