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;
inline int read() {
  int x = 0, f = 1;
  char g = getchar();
  while (g < '0' || g > '9') {
    if (g == '-') f = -1;
    g = getchar();
  }
  while (g >= '0' && g <= '9') {
    x = (x << 1) + (x << 3) + g - 48;
    g = getchar();
  }
  return f * x;
}
int n, q[200005], ne[200005];
long double k[200005];
bool f[200005];
struct node {
  int x, y, id;
  inline bool operator<(node a) const {
    return x > a.x || (a.x == x && y > a.y);
  }
} p[200005];
inline long double slay(node& a, node& b) {
  return (long double)a.x * b.x * (b.y - a.y) /
         ((long double)a.y * b.y * (b.x - a.x));
}
int main() {
  n = read();
  int xx, yy = 0, t;
  for (int i = 1; i <= n; i++) {
    p[i].x = read();
    p[i].y = read();
    p[i].id = i;
    if (p[i].y > yy || (p[i].y == yy && p[i].x > xx)) {
      yy = p[i].y;
      xx = p[i].x;
    }
  }
  sort(p + 1, p + n + 1);
  t = 1;
  q[t] = 1;
  for (int i = 2; i <= n && xx <= 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] > slay(p[q[t]], p[i])) --t;
    q[++t] = i;
    k[t] = slay(p[q[t - 1]], p[i]);
  }
  while (t) {
    for (int i = p[q[t]].id; i; i = ne[i]) f[i] = 1;
    t--;
  }
  for (int i = 1; i <= n; i++) {
    if (f[i]) printf("%d ", i);
  }
  printf("\n");
  return 0;
}
