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 
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 int N = 2e5 + 10;
struct Point {
  int id;
  long double x, y;
  Point(double x = 0, double y = 0) : x(x), y(y) {}
};
const long double EPS = 1e-8;
Point operator+(Point A, Point B) { return Point(A.x + B.x, A.y + B.y); }
Point operator-(Point A, Point B) { return Point(A.x - B.x, A.y - B.y); }
Point operator*(Point A, double p) { return Point(A.x * p, A.y * p); }
Point operator/(Point A, double p) { return Point(A.x / p, A.y / p); }
bool operator<(const Point& a, const Point& b) {
  return a.x < b.x || (a.x == b.x && a.y < b.y);
}
int Sgn(long double x) {
  if (fabs(x) < EPS)
    return 0;
  else
    return x < 0 ? -1 : 1;
}
bool operator==(const Point& a, const Point& b) {
  return Sgn(a.x - b.x) == 0 && Sgn(a.y - b.y) == 0;
}
long double Cross(Point A, Point B) {
  return (long double)A.x * B.y - A.y * B.x;
}
int ConvexHull(Point* p, int n, Point* ch) {
  sort(p, p + n);
  int m = 0;
  for (int i = 0; i < n; i++) {
    while (m > 1 && (Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2])) < 0) m--;
    ch[m++] = p[i];
  }
  int k = m;
  for (int i = n - 2; i >= 0; i--) {
    while (m > k && (Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2])) < 0) m--;
    ch[m++] = p[i];
  }
  if (n > 1) m--;
  return m;
}
int n;
Point p[N], ch[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0, x, y; i < n; i++) {
    cin >> x >> y;
    p[i].id = i + 1;
    p[i].x = 1.0 / x;
    p[i].y = 1.0 / y;
  }
  n = ConvexHull(p, n, ch);
  for (int i = 0; i < n; i++) p[i] = ch[i];
  if (n == 1) {
    cout << p[0].id;
    return 0;
  }
  vector<int> ans;
  int i;
  for (i = 0; i < n; i++) {
    if (Sgn(p[i].x - p[i + 1].x)) break;
  }
  ans.push_back(p[i].id);
  for (i = i + 1; i < n; i++) {
    if (Sgn((p[i].y - p[i - 1].y) / (p[i].x - p[i - 1].x)) < 0)
      ans.push_back(p[i].id);
    else
      break;
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << " \n"[i == (int)ans.size() - 1];
  }
  return 0;
}
