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 pt {
  int x, y, id;
  bool operator<(const pt& a) const { return x < a.x || x == a.x && y < a.y; }
  bool operator==(const pt& a) const { return y == a.y && x == a.x; }
};
const int N = 200005;
vector<int> ids[N];
int ans[N], asz = 0;
bool ok[N];
vector<pt> v, up;
pt arr[N];
bool cwe(pt& a, pt& b, pt& c) {
  return 1ll * a.y * b.x * c.x * (c.y - b.y) +
             1ll * a.x * b.y * c.x * (a.y - c.y) +
             1ll * a.x * b.x * c.y * (b.y - a.y) <=
         0;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &arr[i].x, &arr[i].y);
    arr[i].id = i;
  }
  sort(arr, arr + n);
  v.push_back(arr[0]);
  for (int i = 1; i < n; ++i) {
    while (v.size() &&
           (v[v.size() - 1].y < arr[i].y ||
            v[v.size() - 1].y == arr[i].y && v[v.size() - 1].x < arr[i].x))
      v.pop_back();
    if (v.size() && v[v.size() - 1] == arr[i])
      ids[v[v.size() - 1].id].push_back(arr[i].id);
    else
      v.push_back(arr[i]);
  }
  for (int i = 0; i < v.size(); ++i) {
    while (up.size() > 1 && !cwe(up[up.size() - 2], up[up.size() - 1], v[i]))
      up.pop_back();
    up.push_back(v[i]);
  }
  for (int i = 0; i < up.size(); ++i) {
    ans[asz++] = up[i].id;
    for (int j = 0; j < ids[up[i].id].size(); ++j)
      ans[asz++] = ids[up[i].id][j];
  }
  sort(ans, ans + asz);
  for (int i = 0; i < asz; ++i) printf("%d ", ans[i] + 1);
  printf("\n");
  return 0;
}
