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(300003);
int a[MaxN], b[MaxN], S[MaxN], top;
map<pair<int, int>, int> F;
struct node {
  double x, y;
  inline bool operator<(const node &T) const {
    return x == T.x ? y < T.y : x < T.x;
  }
} p[MaxN];
double cross(const node &p1, const node &p2) {
  return (p1.x * p2.x * (p2.y - p1.y)) / (p1.y * p2.y * (p2.x - p1.x));
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i], &b[i]);
    p[i].x = a[i], p[i].y = b[i];
  }
  sort(p + 1, p + 1 + n);
  for (int i = 1; i <= n; i++) {
    if (p[i].x == p[i + 1].x) continue;
    while (top > 1 &&
           cross(p[S[top - 1]], p[S[top]]) < cross(p[S[top - 1]], p[i]))
      --top;
    S[++top] = i;
  }
  for (int i = 1; i <= top; i++)
    if (i == top || cross(p[S[i]], p[S[i + 1]]) < 0)
      F[pair<int, int>(p[S[i]].x, p[S[i]].y)] = 1;
  for (int i = 1; i <= n; i++)
    if (F.count(pair<int, int>(a[i], b[i]))) printf("%d ", i);
  return 0;
}
