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[300006];
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); }
int mult(unsigned long long x1, unsigned long long y1, unsigned long long x2,
         unsigned long long y2, unsigned long long x3, unsigned long long y3) {
  unsigned long long tmp1, tmp2;
  tmp1 = x3 * y2 * (x1 - x2) * (y1 - y3);
  tmp2 = x2 * y3 * (x1 - x3) * (y1 - y2);
  if (tmp1 > tmp2) return 1;
  if (tmp1 == tmp2) return 0;
  return -1;
}
int main() {
  long i, j, bj1, bj2, l, r;
  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 + 1 + n, cmp1);
  bj1 = a[n].x;
  bj2 = a[n].y;
  for (i = n - 1; i; --i) {
    if (a[i].y <= bj2 && !(a[i].x == bj1 && a[i].y == bj2))
      a[i].rk = 0;
    else {
      bj1 = a[i].x;
      bj2 = a[i].y;
    }
  }
  sort(a + 1, a + 1 + n, cmp2);
  while (!a[n].rk) --n;
  sort(a + 1, a + 1 + n, cmp1);
  ans.clear();
  l = 1;
  while (l <= n && a[l].x == a[1].x && a[l].y == a[1].y) {
    ans.insert(a[l].rk);
    ++l;
  }
  r = n;
  while (r && a[r].x == a[n].x && a[r].y == a[n].y) {
    ans.insert(a[r].rk);
    --r;
  }
  if (n > 2) {
    a[l].lx = a[1].x;
    a[l].ly = a[1].y;
    stk.clear();
    stk.push_back(a[1]);
    stk.push_back(a[l]);
    for (i = l + 1; i <= r; ++i) {
      while (!stk.empty() && stk[stk.size() - 1].x == a[i].x &&
             stk[stk.size() - 1].y == a[i].y)
        stk.pop_back();
      while (stk.size() > 1) {
        if (mult(stk[stk.size() - 2].x, stk[stk.size() - 2].y,
                 stk[stk.size() - 1].x, stk[stk.size() - 1].y, a[i].x,
                 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[r].rx = a[n].x;
    a[r].ry = a[n].y;
    stk.clear();
    stk.push_back(a[n]);
    stk.push_back(a[r]);
    for (i = r - 1; i >= l; --i) {
      while (!stk.empty() && stk[stk.size() - 1].x == a[i].x &&
             stk[stk.size() - 1].y == a[i].y)
        stk.pop_back();
      while (stk.size() > 1) {
        if (mult(stk[stk.size() - 2].x, stk[stk.size() - 2].y,
                 stk[stk.size() - 1].x, stk[stk.size() - 1].y, a[i].x,
                 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 (mult(a[i].lx, a[i].ly, a[i].x, a[i].y, a[i].rx, a[i].ry) <= 0)
        ans.insert(a[i].rk);
  }
  set<long>::iterator ia;
  for (ia = ans.begin(); ia != ans.end(); ++ia) printf("%ld ", *ia);
  return 0;
}
