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 q[300005], top, nxt[300005], ok[300005], n;
double k[300005];
struct point {
  int x, y, id;
} p[300005];
bool operator<(const point &lhs, const point &rhs) {
  return (lhs.x > rhs.x) || ((lhs.x == rhs.x) && (lhs.y > rhs.y));
}
double slope(const point &i, const point &j) {
  return 1.0 * i.x * j.x * (j.y - i.y) / (1.0 * i.y * j.y * (j.x - i.x));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  int minx, miny = 0;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].x >> p[i].y;
    p[i].id = i;
    if ((miny < p[i].y) || ((miny == p[i].y) && (minx < p[i].x))) {
      minx = p[i].x;
      miny = p[i].y;
    }
  }
  sort(p + 1, p + n + 1);
  q[top = 1] = 1;
  for (int i = 2; (i <= n) && (minx <= p[i].x); i++) {
    if (p[q[top]].x == p[i].x) {
      if (p[q[top]].y == p[i].y) {
        nxt[p[i].id] = nxt[p[q[top]].id];
        nxt[p[q[top]].id] = p[i].id;
      }
      continue;
    }
    while ((top > 1) && (k[top] > slope(p[q[top]], p[i]))) top--;
    q[++top] = i;
    k[top] = slope(p[q[top - 1]], p[i]);
  }
  for (int i = top; i; i--)
    for (int j = p[q[i]].id; j; j = nxt[j]) ok[j] = 1;
  for (int i = 1; i <= n; i++)
    if (ok[i]) cout << i << " ";
  cout << endl;
  return 0;
}
