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 maxn = 200010;
const double eps = 1e-8;
int dcmp(double x) {
  if (fabs(x) < eps) return 0;
  return x < 0 ? -1 : 1;
}
struct Point {
  int id;
  double x, y;
  bool operator<(const Point &t) const {
    return dcmp(x - t.x) ? x < t.x : y < t.y;
  }
  Point operator-(const Point &t) const { return (Point){0, x - t.x, y - t.y}; }
  double det(const Point &t) const { return x * t.y - y * t.x; }
} p[maxn];
int n, top, que[maxn], ok[maxn], ans[maxn];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%lf%lf", &p[i].x, &p[i].y);
    p[i].x = 10000 / p[i].x;
    p[i].y = 10000 / p[i].y;
    p[i].id = i + 1;
  }
  sort(p, p + n);
  for (int i = 0; i < n; ++i) {
    if (i && !dcmp(p[i].x - p[i - 1].x)) continue;
    while (
        top > 1 &&
        dcmp((p[que[top - 1]] - p[que[top - 2]]).det(p[i] - p[que[top - 2]])) <
            0)
      --top;
    que[top++] = i;
  }
  for (int i = 0; i < top; ++i)
    if (!i || dcmp(p[que[i]].y - p[que[i - 1]].y) < 0) ok[que[i]] = 1;
  for (int i = 0; i < n; ++i) {
    if (i && !dcmp(p[i].x - p[i - 1].x) && !dcmp(p[i].y - p[i - 1].y))
      ok[i] |= ok[i - 1];
    if (ok[i]) ans[p[i].id] = 1;
  }
  for (int i = 1; i <= n; ++i)
    if (ans[i]) printf("%d ", i);
  puts("");
  return 0;
}
