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;
double add(double a, double b) {
  if (abs(a + b) < 1e-10 * (abs(a) + abs(b))) return 0;
  return a + b;
}
struct P {
  double x, y;
  int id;
  P() {}
  P(double x, double y) : x(x), y(y) {}
  P operator+(P p) { return P(add(x, p.x), add(y, p.y)); }
  P operator-(P p) { return P(add(x, -p.x), add(y, -p.y)); }
  P operator*(double d) { return P(x * d, y * d); }
  double dot(P p) { return add(x * p.x, y * p.y); }
  double det(P p) { return add(x * p.y, -y * p.x); }
};
bool cmp_x(const P& p, const P& q) {
  if (fabs(p.x - q.x) > 1e-10) return p.x < q.x;
  return p.y < q.y;
}
int pos[200000 + 10] = {0};
int judge(double x) { return x < -1e-10; }
vector<P> ps;
vector<P> convex_hull(int n) {
  int k = 0;
  sort(ps.begin(), ps.end(), cmp_x);
  vector<P> qs(n * 2);
  for (int i = 0; i < n; i++) {
    if (i >= 1 && fabs(ps[i].x - ps[i - 1].x) < 1e-10 &&
        fabs(ps[i].y - ps[i - 1].y) < 1e-10) {
      pos[i] = pos[i - 1];
      continue;
    }
    pos[i] = i;
    while (k > 1 && judge((qs[k - 1] - qs[k - 2]).det(ps[i] - qs[k - 1]))) {
      k--;
    }
    qs[k++] = ps[i];
  }
  qs.resize(k);
  return qs;
}
int main() {
  int n;
  scanf("%d", &n);
  double maxv = 1 << 30;
  for (int i = 0; i < n; i++) {
    double x, y;
    scanf("%lf%lf", &x, &y);
    P q(100000 / x, 100000 / y);
    q.id = i;
    ps.push_back(q);
    maxv = min(maxv, 100000 / y);
  }
  vector<P> last = convex_hull(n);
  int vis[200000 + 10] = {0};
  for (int i = 0; i < last.size(); i++) {
    vis[last[i].id] = 1;
    if (fabs(last[i].y - maxv) < 1e-10) {
      break;
    }
  }
  for (int i = 0; i < n; i++) {
    if (vis[ps[pos[i]].id]) {
      vis[ps[i].id] = 1;
    }
  }
  for (int i = 0; i < n; i++)
    if (vis[i]) printf("%d ", i + 1);
  printf("\n");
  return 0;
}
