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 
5
10 50
10 50
10 50
10 50
10 50
 2
10000 10000
10000 10000
 3
10000 10000
10000 10000
10000 10000
 3
1000 3000
1500 1500
3000 1000
 43
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 1
1 1
 2
1 1
1 1
 44
78 52
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 3
1000 3000
1499 1499
3000 1000
 5
50 50
49 50
50 49
49 49
50 1
 18
82 38
33 69
33 69
33 69
33 69
33 69
82 38
33 69
33 69
33 69
82 38
33 69
33 69
82 38
82 38
33 69
33 69
33 69
 1
10000 9999
 3
1000 3000
1501 1501
3000 1000
 2
1 2
2 1

1 2 3 4 5
 1 2
 1 2 3
 1 2 3
 1 3 16 24 29 31 32 37 41 42
 1
 1 2
 1
 1 3
 1
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
 1
 1 2 3
 1 2 

#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-20, mul = 1;
const int maxn = 200000 + 10;
int dcmp(double x) {
  if (fabs(x) < eps) return 0;
  return x > 0 ? 1 : -1;
}
struct pt {
  double x, y;
  bool operator<(const pt &rhs) const {
    if (!dcmp(x - rhs.x)) return y < rhs.y;
    return x < rhs.x;
  }
};
pt operator+(const pt &a, const pt &b) { return (pt){a.x + b.x, a.y + b.y}; }
pt operator-(const pt &a, const pt &b) { return (pt){a.x - b.x, a.y - b.y}; }
pt operator*(const pt &a, const double &d) { return (pt){d * a.x, d * a.y}; }
bool operator==(const pt &a, const pt &b) {
  return !dcmp(a.x - b.x) && !dcmp(a.y - b.y);
}
double dot(const pt &a, const pt &b) { return a.x * b.x + a.y * b.y; }
double cross(const pt &a, const pt &b) { return a.x * b.y - a.y * b.x; }
double cross(const pt &a, const pt &b, const pt &c) {
  return cross(b - a, c - a);
}
pt st[maxn];
void convex_hull(vector<pt> &v, vector<pt> &p) {
  p.clear();
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  int n = v.size();
  int sz = 0;
  for (int i = 0; i < n; i++) {
    while (sz >= 2 && dcmp(cross(st[sz - 2], st[sz - 1], v[i])) < 0) sz--;
    st[sz++] = v[i];
  }
  for (int i = 0; i < sz; i++) p.push_back(st[i]);
  sz = 0;
  for (int i = 0; i < n; i++) {
    while (sz >= 2 && dcmp(cross(st[sz - 2], st[sz - 1], v[i]) > 0)) sz--;
    st[sz++] = v[i];
  }
  for (int i = sz - 1; i >= 0; i--) {
    if (i == sz - 1 && st[i] == p.back()) continue;
    if (i == 0 && st[i] == p[0]) continue;
    p.push_back(st[i]);
  }
}
vector<pt> a, b;
set<pair<int, int> > ans;
int x[maxn], y[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x[i], &y[i]);
    a.push_back((pt){mul / x[i], mul / y[i]});
  }
  convex_hull(a, b);
  for (int i = 0; i < b.size(); i++) {
    if (i && dcmp(b[i - 1].y - b[i].y) <= 0) break;
    int x0 = int(mul / b[i].x + 0.5), y0 = int(mul / b[i].y + 0.5);
    ans.insert((pair<int, int>){x0, y0});
  }
  for (int i = 1; i <= n; i++)
    if (ans.count((pair<int, int>){x[i], y[i]})) printf("%d ", i);
  printf("\n");
}
