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;
void RI() {}
template <typename... T>
void RI(int& head, T&... tail) {
  scanf("%d", &head);
  RI(tail...);
}
int n, s[200010], r[200010], ss[200010], stk[200010], top;
pair<int, int> get(int i) { return make_pair(s[i], r[i]); }
bool cmp(int a, int b) { return get(a) > get(b); }
int main() {
  RI(n);
  for (int i = 0; i < int(n); i++) RI(s[i], r[i]);
  for (int i = 0; i < int(n); i++) ss[i] = i;
  sort(ss, ss + n, cmp);
  for (int ii = 0; ii < int(n); ii++) {
    int i = ss[ii];
    if (ii > 0 && get(i) == get(ss[ii - 1])) continue;
    while (top >= 2) {
      long long s1 = s[i], r1 = r[i];
      long long s2 = s[stk[top - 1]], r2 = r[stk[top - 1]];
      long long s3 = s[stk[top - 2]], r3 = r[stk[top - 2]];
      if (r1 <= r2) break;
      long long u1 = (s2 - s1) * r1, d1 = (r1 - r2) * s1;
      long long u2 = (s3 - s2) * r3, d2 = (r2 - r3) * s3;
      assert(u1 >= 0 && d1 > 0 && u2 >= 0 && d2 > 0);
      if (u1 * d2 >= u2 * d1) break;
      top--;
    }
    if (top == 0 || (top > 0 && r[i] > r[stk[top - 1]])) stk[top++] = i;
  }
  set<pair<int, int> > st;
  for (int i = 0; i < int(top); i++) st.insert(get(stk[i]));
  vector<int> sol;
  for (int i = 0; i < int(n); i++)
    if (st.count(get(i))) sol.push_back(i);
  for (int i = 0; i < int(((int)(sol).size())); i++)
    printf("%d%c", sol[i] + 1, i + 1 == ((int)(sol).size()) ? '\n' : ' ');
  return 0;
}
