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;
const int N = 3e5 + 9;
int q[N], ne[N];
double k[N];
bool f[N];
char buf[N], *ie = buf + N, *ip = ie - 1;
struct Point {
  int x, y, id;
  inline bool operator<(register Point a) const {
    return x > a.x || (x == a.x && y > a.y);
  }
} p[N];
inline int read() {
  int res = 0, fh = 1;
  char ch = getchar();
  while ((ch > '9' || ch < '0') && ch != '-') ch = getchar();
  if (ch == '-') fh = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') res = res * 10 + ch - '0', ch = getchar();
  return fh * res;
}
inline double Slope(register Point& i, register Point& j) {
  return (double)i.x * j.x * (j.y - i.y) / ((double)i.y * j.y * (j.x - i.x));
}
int main() {
  int n = read(), t, ry = 0, rx;
  for (int i = 1; i <= n; ++i) {
    p[i].x = read();
    p[i].y = read();
    p[i].id = i;
    if (ry < p[i].y || (ry == p[i].y && rx < p[i].x)) ry = p[i].y, rx = p[i].x;
  }
  sort(p + 1, p + n + 1);
  q[t = 1] = 1;
  for (int i = 2; i <= n && rx <= p[i].x; ++i) {
    if (p[q[t]].x == p[i].x) {
      if (p[q[t]].y == p[i].y)
        ne[p[i].id] = ne[p[q[t]].id], ne[p[q[t]].id] = p[i].id;
      continue;
    }
    while (t > 1 && k[t] > Slope(p[q[t]], p[i])) --t;
    q[++t] = i;
    k[t] = Slope(p[q[t - 1]], p[i]);
  }
  for (; t; --t)
    for (int i = p[q[t]].id; i; i = ne[i]) f[i] = 1;
  for (int i = 1; i <= n; ++i)
    if (f[i]) printf("%d ", i);
  puts("");
  return 0;
}
