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 = 1e5 * 2 + 5;
long long Abs(long long x) { return x > 0 ? x : -x; }
long long gcd(long long x, long long y) {
  x = Abs(x);
  y = Abs(y);
  return y ? gcd(y, x % y) : x;
}
struct fen {
  long long i, j;
  fen() {}
  fen(long long i, long long j) : i(i), j(j) {}
  fen operator+(const fen &c) const {
    long long a = i * c.j + j * c.i;
    long long b = j * c.j;
    long long g = Abs(gcd(a, b));
    return fen(a / g, b / g);
  }
  fen operator-(const fen &c) const {
    long long a = i * c.j - j * c.i;
    long long b = j * c.j;
    long long g = Abs(gcd(a, b));
    return fen(a / g, b / g);
  }
  fen operator*(const fen &c) const {
    long long a = i * c.i;
    long long b = j * c.j;
    long long g = Abs(gcd(a, b));
    return fen(a / g, b / g);
  }
  bool operator<(const fen &c) const { return j > c.j; }
  bool operator==(const fen &c) const { return i == c.i && c.j == j; }
};
struct point {
  vector<int> id;
  fen x, y;
  bool operator<(const point &c) const {
    return x < c.x || (x == c.x && y < c.y);
  }
  void chu(point c) {
    x = c.x;
    y = c.y;
    id.clear();
    for (int j = c.id.size() - 1; j >= 0; j--) id.push_back(c.id[j]);
  }
} ps[N], qs[N];
fen cha(point a, point b, point c) {
  point p, q;
  p.x = (a.x - b.x);
  p.y = (a.y - b.y);
  q.x = (c.x - a.x);
  q.y = (c.y - a.y);
  return (p.x * q.y) - (p.y * q.x);
}
vector<int> out;
int main() {
  int n;
  while (cin >> n) {
    for (int i = 0; i < n; i++) {
      ps[i].id.clear();
      ps[i].id.push_back(i + 1);
      int a, b;
      scanf("%d%d", &a, &b);
      ps[i].x.j = a, ps[i].y.j = b;
      ps[i].x.i = ps[i].y.i = 1;
    }
    sort(ps, ps + n);
    int k = 1;
    fen ma = ps[0].y;
    qs[0].chu(ps[0]);
    for (int i = 1; i < n; i++) {
      if (qs[k - 1].x == ps[i].x && qs[k - 1].y == ps[i].y) {
        qs[k - 1].id.push_back(ps[i].id[0]);
        continue;
      }
      if (ps[i].y < ma) {
        while (k > 1 && cha(qs[k - 1], qs[k - 2], ps[i]) < fen(0, 1e9 + 7)) {
          k--;
        }
        qs[k++].chu(ps[i]);
        ma = qs[k - 1].y;
      }
    }
    out.clear();
    for (int i = 0; i < k; i++)
      for (int j = qs[i].id.size() - 1; j >= 0; j--) out.push_back(qs[i].id[j]);
    sort(out.begin(), out.end());
    printf("%d", out[0]);
    n = out.size();
    for (int i = 1; i < n; i++) printf(" %d", out[i]);
    printf("\n");
  }
  return 0;
}
