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 ans[1111111], U, tans[1111111];
int N;
map<pair<int, int>, int> m;
int M = 0;
vector<int> v[1111111];
pair<int, int> c[1111111];
long double s(int A, int B, int C) {
  return (((long double)1.0) / c[B].first - ((long double)1.0) / c[A].first) *
             (((long double)1.0) / c[C].second -
              ((long double)1.0) / c[A].second) -
         (((long double)1.0) / c[B].second - ((long double)1.0) / c[A].second) *
             (((long double)1.0) / c[C].first -
              ((long double)1.0) / c[A].first);
}
void convex_hull() {
  sort(c, c + N);
  U = 0;
  for (int i = 0; i < (N); i++) {
    while ((U && c[tans[U - 1]].second <= c[i].second) ||
           (U > 1 && s(i, tans[U - 1], tans[U - 2]) < -1e-15))
      U--;
    tans[U] = i;
    ans[U++] = m[c[i]];
  }
}
vector<int> t;
int main(int argc, char *argv[]) {
  int M;
  scanf("%d", &M);
  for (int i = 0; i < (M); i++) {
    scanf("%d%d", &c[N].first, &c[N].second);
    pair<int, int> p = c[N];
    if (!m.count(p)) m[p] = N++;
    v[m[p]].push_back(i);
  }
  convex_hull();
  for (int i = 0; i < (U); i++)
    t.insert(t.end(), v[ans[i]].begin(), v[ans[i]].end());
  sort(t.begin(), t.end());
  for (int i = 0; i < (t.size()); i++) {
    if (i) printf(" ");
    printf("%d", t[i] + 1);
  }
  printf("\n");
  return 0;
}
