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;
const int maxn = 200005;
const double eps = 1e-9;
int n, b[maxn], c[maxn], stak[maxn], top, ans[maxn], cnt;
pair<int, int> p[maxn];
vector<int> G[maxn];
struct node {
  double fir, sec;
  node() {}
  node(double x, double y) {
    fir = x;
    sec = y;
  }
} a[maxn];
node operator-(node x, node y) { return node(x.fir - y.fir, x.sec - y.sec); }
double crp(node x, node y) { return x.fir * y.sec - x.sec * y.fir; }
bool cmp(int x, int y) {
  return fabs(a[x].fir - a[y].fir) < eps ? a[x].sec > a[y].sec
                                         : a[x].fir < a[y].fir;
}
bool cmp1(int x, int y) { return p[x] < p[y]; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &p[i].first, &p[i].second);
    b[i] = i;
  }
  sort(b + 1, b + n + 1, cmp1);
  int m = 0;
  for (int l = 1, r; l <= n; l = r + 1) {
    r = l;
    while (r < n && p[b[l]] == p[b[r + 1]]) ++r;
    if (r < n && p[b[r]].first == p[b[r + 1]].first) continue;
    ++m;
    c[m] = m;
    a[m] = node(1.0 / p[b[l]].first, 1.0 / p[b[l]].second);
    for (int j = l; j <= r; ++j) G[m].push_back(b[j]);
  }
  sort(c + 1, c + m + 1, cmp);
  for (int i = 1; i <= m; ++i) {
    while (top > 1 &&
           crp(a[stak[top]] - a[stak[top - 1]], a[c[i]] - a[stak[top - 1]]) < 0)
      --top;
    stak[++top] = c[i];
  }
  for (int i = 0; i < G[stak[1]].size(); ++i) ans[++cnt] = G[stak[1]][i];
  for (int i = 2; i <= top; ++i) {
    node u = a[stak[i]] - a[stak[i - 1]];
    if (u.sec < 0) {
      for (int j = 0; j < G[stak[i]].size(); ++i) ans[++cnt] = G[stak[i]][j];
    } else {
      break;
    }
  }
  sort(ans + 1, ans + cnt + 1);
  for (int i = 1; i <= cnt; ++i) printf("%d ", ans[i]);
  return 0;
}
