Tavas is a cheerleader in the new sports competition named "Pashmaks".

<image>

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;
pair<pair<int, int>, int> P[200020];
int MX, a, b, n, i, j;
vector<int> ans, all[200020];
double L, R, S[12000], T[12000], X[200020], Y[200020];
bool f[200020];
int main() {
  cin >> n;
  int NN = n;
  for (i = 1; i <= n; i++) {
    scanf("%d%d", &a, &b);
    X[i] = a;
    Y[i] = b;
    P[i].first = make_pair(a, b);
    P[i].second = i;
  }
  sort(P + 1, P + n + 1);
  reverse(P + 1, P + n + 1);
  for (i = 1; i <= n; i++) {
    j = i;
    while (j <= n && P[j].first == P[i].first) j++;
    for (int k = i; k < j; k++) {
      all[P[i].second].push_back(P[k].second);
    }
    if (MX < P[i].first.second) ans.push_back(P[i].second);
    i = j - 1;
    for (int k = i; k < j; k++) MX = max(MX, P[k].first.second);
  }
  n = 0;
  sort(ans.begin(), ans.end());
  for (j = 0; j < ans.size(); j++) {
    ++n;
    P[n].first.first = X[ans[j]];
    P[n].first.second = Y[ans[j]];
    P[n].second = ans[j];
  }
  sort(P + 1, P + n + 1);
  for (i = 1; i <= n; i++) {
    T[i] = P[i].first.second;
    S[i] = P[i].first.first;
  }
  for (i = 1; i <= n; i++) {
    L = 0;
    R = 1000000000000;
    for (j = 1; j < i; j++)
      L = ((L) > ((T[j] - T[i]) * S[j] / ((S[i] - S[j]) * T[j]))
               ? (L)
               : ((T[j] - T[i]) * S[j] / ((S[i] - S[j]) * T[j])));
    for (j = i + 1; j <= n; j++)
      R = ((R) < ((T[i] - T[j]) * S[j] / ((S[j] - S[i]) * T[j]))
               ? (R)
               : ((T[i] - T[j]) * S[j] / ((S[j] - S[i]) * T[j])));
    if (L <= R) f[P[i].second] = 1;
  }
  vector<int> res;
  for (i = 1; i <= NN; i++)
    if (f[i]) {
      for (j = 0; j < all[i].size(); j++) res.push_back(all[i][j]);
    }
  sort(res.begin(), res.end());
  for (j = 0; j < res.size(); j++) cout << res[j] << " ";
}
