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;
inline void OPEN(const string &s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
double _acos(double x) {
  double ret = acos(x);
  if (ret == ret) return ret;
  if (x < 0) return acos(-1.0);
  return acos(1.0);
}
const double PI = _acos(-1);
const double EPS = 1e-50;
const double INF = 1e6 + 100;
struct point {
  double x, y;
  int id;
  point() {
    x = y = 0.;
    id = 0;
  }
  point(double x, double y) : x(x), y(y) {}
  point(double x, double y, int id) : x(x), y(y), id(id) {}
};
point operator-(const point &p1, const point &p2) {
  return point(p1.x - p2.x, p1.y - p2.y);
}
point operator+(const point &p1, const point &p2) {
  return point(p1.x + p2.x, p1.y + p2.y);
}
point operator*(const point &p, const double &s) {
  return point(p.x * s, p.y * s);
}
point operator/(const point &p, const double &s) { return p * (1. / s); }
bool operator<(const point &p1, const point &p2) {
  if (p1.x != p2.x) return p1.x < p2.x;
  return p1.y < p2.y;
}
bool operator==(const point &p1, const point &p2) {
  return fabs(p1.x - p2.x) < EPS && fabs(p1.y - p2.y) < EPS;
}
double cross(const point &p1, const point &p2) {
  return p1.x * p2.y - p1.y * p2.x;
}
bool isRightTurn(const point &p1, const point &p2, const point &p3) {
  return cross(p2 - p1, p3 - p2) <= 0.;
}
vector<point> convexHull(vector<point> p) {
  int m = 0, n = p.size();
  vector<point> hull(2 * n);
  sort(p.begin(), p.end());
  for (int i = 0; i < n; ++i) {
    while (m >= 2 && isRightTurn(hull[m - 2], hull[m - 1], p[i])) --m;
    hull[m++] = p[i];
  }
  for (int i = n - 1, t = m + 1; i >= 0; --i) {
    while (m >= t && isRightTurn(hull[m - 2], hull[m - 1], p[i])) --m;
    hull[m++] = p[i];
  }
  hull.resize(m);
  return hull;
}
int n;
vector<point> poly;
int ans[200005];
map<point, int> h;
int group[200005];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) {
    double r, s;
    scanf("%lf %lf", &r, &s);
    if (h.count(point(r, s)) == 0) h[point(r, s)] = i;
    group[i] = h[point(r, s)];
    poly.push_back(point(1. / r, 1. / s, i));
  }
  vector<point> hull = convexHull(poly);
  double lo = INF;
  for (int i = 0; i < (((int)(hull).size())); ++i) {
    if (EPS < lo - hull[i].y) {
      lo = hull[i].y;
      ans[group[hull[i].id]] = 1;
    }
  }
  int flag = 0;
  for (int i = 0; i < (n); ++i) {
    if (ans[group[i]] == 0) continue;
    if (flag)
      printf(" ");
    else
      flag = 1;
    printf("%d", i + 1);
  }
  printf("\n");
  return 0;
}
