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;
struct Node {
  int x, y, idx;
} p[200005];
bool che(Node A, Node B, Node C) {
  return 1ll * (C.x - B.x) * (A.y - B.y) * C.y * A.x <
         1ll * (A.x - B.x) * (C.y - B.y) * A.y * C.x;
}
Node v[200005];
bool cmp(Node A, Node B) { return A.x < B.x || A.x == B.x && A.y < B.y; }
int n;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &p[i].x, &p[i].y);
    p[i].idx = i;
  }
  sort(p + 1, p + n + 1, cmp);
  int tail = 0;
  for (int i = 1; i <= n; i++) {
    while (1 <= tail && v[tail].y <= p[i].y) tail--;
    while (1 < tail && che(p[i], v[tail], v[tail - 1])) tail--;
    v[++tail] = p[i];
  }
  set<pair<int, int> > q;
  for (int i = 1; i <= tail; i++) q.insert(make_pair(v[i].x, v[i].y));
  vector<int> ans;
  for (int i = 1; i <= n; i++)
    if (q.count(make_pair(p[i].x, p[i].y))) ans.push_back(p[i].idx);
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  return 0;
}
