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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int maxn = 200010;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
double add(double a, double b) {
  if (abs(a + b) < eps * (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) > eps) return p.x < q.x;
  return p.y < q.y;
}
int fa[maxn], vis[maxn];
P ps[maxn];
vector<P> convex_hull(P* ps, int n) {
  int k = 0;
  sort(ps, ps + n, 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) < eps &&
        fabs(ps[i].y - ps[i - 1].y) < eps) {
      fa[i] = fa[i - 1];
      continue;
    }
    fa[i] = i;
    while (k > 1 && ((qs[k - 1] - qs[k - 2]).det(ps[i] - qs[k - 1]) < 0)) k--;
    qs[k++] = ps[i];
  }
  return qs;
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    double my = inf;
    for (int i = 0; i < n; i++) {
      double x, y;
      scanf("%lf%lf", &x, &y);
      P p(1000 / x, 1000 / y);
      p.id = i;
      ps[i] = p;
      my = min(my, 1000 / y);
    }
    vector<P> qs = convex_hull(ps, n);
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < qs.size(); i++) {
      vis[qs[i].id] = 1;
      if (fabs(my - qs[i].y) < eps) break;
    }
    for (int i = 0; i < n; i++) {
      if (vis[ps[fa[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;
}
