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;
int pd[210000], s[210000], head, n;
struct atom {
  int where, x, y;
} x[210000];
int compre(atom k1, atom k2) {
  return k1.x < k2.x || (k1.x == k2.x && k1.y > k2.y);
}
int check(int i) {
  int k1 = x[s[head - 1]].y, k2 = x[s[head]].y, k3 = x[i].y,
      k4 = x[s[head - 1]].x, k5 = x[s[head]].x, k6 = x[i].x;
  return 1ll * (k1 * k3 - k2 * k3) * (k4 * k6 - k4 * k5) >
         1ll * (k1 * k2 - k1 * k3) * (k5 * k6 - k4 * k6);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &x[i].x, &x[i].y);
    x[i].where = i;
  }
  sort(x + 1, x + 1 + n, compre);
  head = 1;
  s[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (x[i].x == x[i - 1].x) continue;
    while (head >= 2 && check(i)) head--;
    s[++head] = i;
  }
  for (int i = 1; i <= head; i++) pd[x[s[i]].where] = 1;
  for (int i = 2; i <= head; i++)
    if (x[s[i - 1]].y <= x[s[i]].y)
      pd[x[s[i - 1]].where] = 0;
    else
      break;
  for (int i = 2; i <= n; i++) {
    if (x[i].y == x[i - 1].y && x[i].x == x[i - 1].x && pd[x[i - 1].where])
      pd[x[i].where] = 1;
  }
  for (int i = 1; i <= n; i++)
    if (pd[i]) printf("%d ", i);
  cout << endl;
  return 0;
}
