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 EE {
  int s, r, index;
};
EE a[200002] = {{0}}, b[200002] = {{0}}, c[200002] = {{0}};
int n, fir = 1, las = 1, bl = 0, cl = 0;
long long fm[200002] = {0}, sfz[200002] = {0}, rfz[200002] = {0}, nowfm = 0,
          nowsfz = 0, nowrfz = 0;
long double zhi[200002] = {0};
bool youjd = 0;
int comp(EE xx, EE yy) {
  if (xx.s == yy.s && xx.r == yy.r) return 0;
  if (xx.s >= yy.s && xx.r >= yy.r) return 1;
  if (xx.s <= yy.s && xx.r <= yy.r) return -1;
  return 0;
}
bool comp0(EE xx, EE yy) {
  return (xx.s > yy.s || xx.s == yy.s && xx.r > yy.r);
}
bool comp1(EE xx, EE yy) { return (xx.index < yy.index); }
long long hls(long long aa, long long bb, long long cc, long long dd) {
  return aa * dd - bb * cc;
}
int main() {
  int i, cc;
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
  cin >> n;
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].s, &a[i].r);
    a[i].index = i;
  }
  sort(a + 1, a + n + 1, comp0);
  bl = 1;
  b[1].s = a[1].s;
  b[1].r = a[1].r;
  b[1].index = a[1].index;
  for (i = 2; i <= n; i++) {
    if (a[i].r > b[bl].r || (a[i].s == b[bl].s && a[i].r == b[bl].r)) {
      bl++;
      b[bl].s = a[i].s;
      b[bl].r = a[i].r;
      b[bl].index = a[i].index;
    }
  }
  cl = 1;
  c[1].s = b[1].s;
  c[1].r = b[1].r;
  c[1].index = b[1].index;
  for (i = 2; i <= bl; i++) {
    if (c[cl].s == b[i].s && c[cl].r == b[i].r) {
      cl++;
      c[cl].s = b[i].s;
      c[cl].r = b[i].r;
      c[cl].index = b[i].index;
      fm[cl] = fm[cl - 1];
      sfz[cl] = sfz[cl - 1];
      rfz[cl] = rfz[cl - 1];
      zhi[cl] = zhi[cl - 1];
    } else {
      nowfm = hls(c[cl].r, c[cl].s, b[i].r, b[i].s);
      nowsfz = hls(c[cl].r, c[cl].s * c[cl].r, b[i].r, b[i].s * b[i].r);
      nowrfz = hls(c[cl].s, c[cl].s * c[cl].r, b[i].s, b[i].s * b[i].r);
      if (fm[cl] == 0) {
        cl++;
        c[cl].s = b[i].s;
        c[cl].r = b[i].r;
        c[cl].index = b[i].index;
        fm[cl] = nowfm;
        sfz[cl] = nowsfz;
        rfz[cl] = nowrfz;
        zhi[cl] = nowsfz * 1.0 / nowfm;
      } else {
        while (nowsfz * 1.0 / nowfm - zhi[cl] < -1e-6) {
          c[cl].s = 0;
          c[cl].r = 0;
          c[cl].index = 0;
          fm[cl] = 0;
          sfz[cl] = 0;
          rfz[cl] = 0;
          cl--;
          if (fm[cl] == 0) break;
        }
        cl++;
        c[cl].s = b[i].s;
        c[cl].r = b[i].r;
        c[cl].index = b[i].index;
        fm[cl] = nowfm;
        sfz[cl] = nowsfz;
        rfz[cl] = nowrfz;
        zhi[cl] = nowsfz * 1.0 / nowfm;
      }
    }
  }
  sort(c + 1, c + cl + 1, comp1);
  for (i = 1; i <= cl; i++)
    if (i == cl)
      printf("%d\n", c[i].index);
    else
      printf("%d ", c[i].index);
  return 0;
}
