Tavas is a cheerleader in the new sports competition named "Pashmaks".

<image>

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 Nmax = 200005, Vmax = 10005;
const double eps = 1e-10;
struct Point {
  int x, y;
  int pos;
  bool operator<(const Point& other) const {
    if (x != other.x) return x > other.x;
    return y > other.y;
  }
  bool operator==(const Point& other) const {
    return x == other.x && y == other.y;
  }
};
Point A[Nmax], Hull[Nmax];
int dist(const Point& a, const Point& b) {
  return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
Point O;
double slope(const Point& a, const Point& b) {
  if (a.x == b.x) return 1e99;
  return ((double)(a.y - b.y) * a.x * b.x) / ((double)(a.x - b.x) * a.y * b.y);
}
struct cmp {
  bool operator()(const Point& a, const Point& b) const {
    double s1 = slope(O, a), s2 = slope(O, b);
    if (abs(s1 - s2) < eps) return dist(O, a) < dist(O, b);
    return s1 + eps < s2;
  }
};
long long crossProduct(const Point& o, const Point& a, const Point& b) {
  return (1LL * a.y * b.x * o.x * o.y - 1LL * a.y * o.x * b.x * b.y -
          1LL * o.y * b.x * a.x * a.y + 1LL * a.x * b.x * a.y * b.y) -
         (1LL * a.x * b.y * o.x * o.y - 1LL * a.x * o.y * b.x * b.y -
          1LL * o.x * b.y * a.x * a.y + 1LL * a.x * b.x * a.y * b.y);
}
int convexHull(int N) {
  int pos = 1;
  for (int i = 2; i <= N; ++i)
    if (A[i] < A[pos]) pos = i;
  swap(A[1], A[pos]);
  O = A[1];
  sort(A + 2, A + N + 1, cmp());
  int maxy = 0;
  for (int i = 1; i <= N; ++i) maxy = max(maxy, A[i].y);
  pos = 1;
  Hull[1] = A[1];
  if (Hull[1].y == maxy) return 1;
  for (int i = 2; i <= N; ++i) {
    if (A[i] == Hull[pos]) continue;
    while (pos >= 2 && crossProduct(Hull[pos - 1], Hull[pos], A[i]) < 0) --pos;
    Hull[++pos] = A[i];
    if (A[i].y == maxy) return pos;
  }
  return pos;
}
int main() {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    cin >> A[i].x >> A[i].y;
    A[i].pos = i;
  }
  int M = convexHull(N);
  vector<int> ans;
  set<Point> hull;
  for (int i = 1; i <= M; ++i) hull.insert(Hull[i]);
  for (int i = 1; i <= N; ++i)
    if (hull.count(A[i]) != 0) ans.push_back(A[i].pos);
  sort(ans.begin(), ans.end());
  for (int p : ans) cout << p << ' ';
  cout << endl;
}
