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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
struct dv {
  long long so, mo;
  dv() {}
  dv(long long so, long long mo) {
    this->so = so;
    this->mo = mo;
  }
  dv operator-(dv a) const {
    dv t;
    t.mo = lcm(mo, a.mo);
    t.so = t.mo / mo * so - t.mo / a.mo * a.so;
    long long tt = gcd(abs(t.mo), abs(t.so));
    t.mo /= tt;
    t.so /= tt;
    return t;
  }
  dv operator*(dv a) const {
    dv t;
    t.so = a.so * so;
    t.mo = a.mo * mo;
    long long tt = gcd(abs(t.so), abs(t.mo));
    t.mo /= tt;
    t.so /= tt;
    return t;
  }
};
struct point {
  int id;
  dv x, y;
  point() {}
  point(dv x, dv y) {
    this->x = x;
    this->y = y;
  }
  point operator-(point a) { return point(x - a.x, y - a.y); }
  dv operator/(point a) { return x * a.y - y * a.x; }
  bool operator<(point a) const {
    return x.mo != a.x.mo ? x.mo > a.x.mo : y.mo < a.y.mo;
  }
};
int covexhull(point p[], int n, point ch[]) {
  sort(p, p + n);
  int m = 0;
  for (int i = 0; i < n; i++) {
    if (m > 0 && p[i].y.mo < ch[m - 1].y.mo) continue;
    while (m > 1 && ((ch[m - 1] - ch[m - 2]) / (p[i] - ch[m - 2])).so < 0) m--;
    ch[m++] = p[i];
  }
  return m;
}
point bx[200010], ch[200010];
vector<int> ans;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    bx[i].id = i;
    bx[i].x.so = bx[i].y.so = 1;
    bx[i].x.mo = a;
    bx[i].y.mo = b;
  }
  n = covexhull(bx, n, ch);
  for (int i = 0; i < n; i++) ans.push_back(ch[i].id);
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i] + 1);
}
