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;
map<pair<long long, long long>, int> mp;
int n, top, q[200010], tt;
long long mx;
struct Vec {
  long long x0, x1, y0, y1;
  int id;
} a[200010];
inline int Read() {
  char c = getchar();
  int num = 0;
  while ('0' > c || c > '9') c = getchar();
  while ('0' <= c && c <= '9') num = num * 10 + c - '0', c = getchar();
  return (num);
}
bool Cmp(Vec a, Vec b) {
  return ((a.x1 > b.x1) || (a.x1 == b.x1 && a.y1 > b.y1));
}
long double operator*(Vec a, Vec b) {
  return ((long double)a.x0 * b.y0 * a.y1 * b.x1 -
          (long double)a.y0 * b.x0 * a.x1 * b.y1);
}
Vec operator-(Vec a, Vec b) {
  return ((Vec){a.x0 * b.x1 - a.x1 * b.x0, a.x1 * b.x1,
                a.y0 * b.y1 - a.y1 * b.y0, a.y1 * b.y1});
}
int main() {
  n = Read();
  for (int i = 1; i <= n; i++)
    a[i].x0 = a[i].y0 = 1, a[i].x1 = Read(), a[i].y1 = Read(), a[i].id = i;
  sort(a + 1, a + 1 + n, Cmp);
  q[top = 1] = 1;
  for (int i = 2; i <= n; i++)
    if (a[i].x1 != a[i - 1].x1) {
      while (top > 1 && (a[i] - a[q[top]]) * (a[q[top]] - a[q[top - 1]]) > 1e-7)
        top--;
      q[++top] = i;
    }
  for (int i = 1; i <= top; i++)
    if (a[q[i]].y1 > mx) tt = i, mx = a[q[i]].y1;
  top = tt;
  for (int i = 1; i <= top; i++) mp[make_pair(a[q[i]].x1, a[q[i]].y1)] = 1;
  top = 0;
  for (int i = 1; i <= n; i++)
    if (mp[make_pair(a[i].x1, a[i].y1)]) q[++top] = a[i].id;
  sort(q + 1, q + 1 + top);
  for (int i = 1; i <= top; i++) printf("%d ", q[i]);
  puts("");
}
