Tavas is a cheerleader in the new sports competition named "Pashmaks".

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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 
#include <bits/stdc++.h>
using namespace std;
struct node {
  int s, r;
  int id;
} a[200050], st[200050], b[200050];
bool cmp(node a, node b) { return a.s > b.s || (a.s == b.s && a.r > b.r); }
int sig(double x) { return x > 1e-8 ? 1 : x < -1e-8 ? -1 : 0; }
bool check(node a, node b, node c) {
  double k1 = (1.0 / c.r - 1.0 / b.r) / (1.0 / b.s - 1.0 / c.s);
  double k2 = (1.0 / b.r - 1.0 / a.r) / (1.0 / a.s - 1.0 / b.s);
  if (sig(k1 - k2) <= 0)
    return true;
  else
    return false;
}
int q[200050];
map<pair<int, int>, int> mp;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &a[i].s, &a[i].r);
    a[i].id = i;
    b[i] = a[i];
  }
  sort(a, a + n, cmp);
  int mx = a[0].r, tail = 1;
  for (int i = 1; i < n; ++i) {
    if (a[i].r <= mx) continue;
    mx = a[i].r;
    a[tail++] = a[i];
  }
  int top = 0;
  for (int i = 0; i < tail; ++i) {
    if (top < 2) {
      st[top++] = a[i];
    } else {
      while (top >= 2 && !check(st[top - 2], st[top - 1], a[i])) top--;
      st[top++] = a[i];
    }
  }
  tail = 0;
  for (int i = 0; i < top; ++i) {
    int x = st[i].id;
    mp[make_pair(b[x].s, b[x].r)] = 1;
  }
  for (int i = 0; i < n; ++i) {
    if (mp.find(make_pair(b[i].s, b[i].r)) != mp.end()) q[tail++] = i;
  }
  sort(q, q + tail);
  for (int i = 0; i < tail; ++i) {
    printf("%d", q[i] + 1);
    if (i == tail - 1)
      puts("");
    else
      printf(" ");
  }
}
