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;
struct point {
  long x, y, rk, lx, ly, rx, ry;
} a[200005];
long n;
set<long> ans;
vector<point> stk;
bool cmp1(point a, point b) { return (a.x < b.x || (a.x == b.x && a.y < b.y)); }
bool cmp2(point a, point b) { return (a.rk > b.rk); }
template <class T>
T mul(T x1, T y1, T x2, T y2, T x3, T y3) {
  return ((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1));
}
int main() {
  long i, j, bj;
  scanf("%ld", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%ld%ld", &a[i].x, &a[i].y);
    a[i].rk = i;
  }
  sort(a + 1, a + n + 1, cmp1);
  bj = a[n].y;
  for (i = n - 1; i; --i) {
    if (a[i].y <= bj)
      a[i].rk = 0;
    else
      bj = a[i].y;
  }
  sort(a + 1, a + n + 1, cmp2);
  while (!a[n].rk) --n;
  sort(a + 1, a + n + 1, cmp1);
  ans.clear();
  ans.insert(a[1].rk);
  ans.insert(a[n].rk);
  if (n > 2) {
    a[2].lx = a[1].x;
    a[2].ly = a[1].y;
    stk.clear();
    stk.push_back(a[1]);
    stk.push_back(a[2]);
    for (i = 3; i < n; ++i) {
      while (stk.size() > 1) {
        if (mul(1 / (double)stk[stk.size() - 2].x,
                1 / (double)stk[stk.size() - 2].y,
                1 / (double)stk[stk.size() - 1].x,
                1 / (double)stk[stk.size() - 1].y, 1 / (double)a[i].x,
                1 / (double)a[i].y) > 0)
          stk.pop_back();
        else
          break;
      }
      a[i].lx = stk[stk.size() - 1].x;
      a[i].ly = stk[stk.size() - 1].y;
      stk.push_back(a[i]);
    }
    a[n - 1].rx = a[n].x;
    a[2].ry = a[n].y;
    stk.clear();
    stk.push_back(a[n]);
    stk.push_back(a[n - 1]);
    for (i = n - 2; i > 1; --i) {
      while (stk.size() > 1) {
        if (mul(1 / (double)stk[stk.size() - 2].x,
                1 / (double)stk[stk.size() - 2].y,
                1 / (double)stk[stk.size() - 1].x,
                1 / (double)stk[stk.size() - 1].y, 1 / (double)a[i].x,
                1 / (double)a[i].y) < 0)
          stk.pop_back();
        else
          break;
      }
      a[i].rx = stk[stk.size() - 1].x;
      a[i].ry = stk[stk.size() - 1].y;
      stk.push_back(a[i]);
    }
    for (i = 2; i < n; ++i)
      if (mul(1 / (double)a[i].lx, 1 / (double)a[i].ly, 1 / (double)a[i].x,
              1 / (double)a[i].y, 1 / (double)a[i].rx, 1 / (double)a[i].ry) < 0)
        ans.insert(a[i].rk);
  }
  set<long>::iterator ia;
  for (ia = ans.begin(); ia != ans.end(); ++ia) printf("%ld ", *ia);
  printf("\n");
  return 0;
}
