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 INF = 1000000003;
int m;
int n;
int x[200001];
int y[200002];
int p[200002];
int l;
int s[200002];
vector<int> v[200002];
bool cmp(int i, int j) { return x[i] > x[j] || (x[i] == x[j] && y[i] > y[j]); }
bool cross(int a, int b, int c) {
  return (long long)(x[a] - x[c]) * (y[a] - y[b]) * x[b] * y[c] >
         (long long)(y[a] - y[c]) * (x[a] - x[b]) * x[c] * y[b];
}
int main(int argc, char *argv[]) {
  cin >> n;
  for (int i = 0; i < n; ++i) scanf("%d %d", x + i, y + i);
  for (int i = 0; i < n; ++i) p[i] = i;
  sort(p, p + n, cmp);
  m = 1;
  s[0] = p[0];
  for (int i = 1; i < n; ++i) {
    int j = p[i];
    if (x[s[m - 1]] == x[j] && y[s[m - 1]] == y[j]) {
      v[s[m - 1]].push_back(j);
      continue;
    }
    if (y[j] <= y[s[m - 1]]) continue;
    while (m > 1 && cross(s[m - 2], s[m - 1], j)) --m;
    s[m++] = j;
  }
  for (int i = 0; i < m; ++i) {
    for (int j : v[s[i]]) s[m++] = j;
  }
  sort(s, s + m);
  for (int i = 0; i < m; ++i) printf("%d ", s[i] + 1);
  cout << endl;
}
