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 nmax = 10000 + 18;
const double EPS = 1e-10;
int S[nmax], LS[nmax], RS[nmax];
vector<int> Q[nmax];
int n;
double A[nmax];
bool ed[200000 + 18];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int s, r;
    scanf("%d%d", &s, &r);
    if (S[s] < r) {
      Q[s].clear();
      S[s] = r;
      Q[s].push_back(i);
    } else if (S[s] == r)
      Q[s].push_back(i);
  }
  for (int i = 1; i <= 10000; ++i)
    if (S[i]) A[i] = i * 1.0 / S[i];
  for (int i = 1; i <= 10000; ++i)
    if (S[i]) {
      double L = -1e100, R = 1e100;
      for (int j = 1; j < i; ++j)
        if (S[j] > S[i]) L = max(L, -A[j] * A[i] * (S[i] - S[j]) / (i - j));
      for (int j = i + 1; j <= 10000; ++j)
        if (S[j] < S[i] && S[j])
          R = min(R, -A[j] * A[i] * (S[i] - S[j]) / (i - j));
        else if (S[j]) {
          R = -1e100;
          break;
        }
      if (L < R + EPS)
        for (int j = 0; j < (int)Q[i].size(); ++j) ed[Q[i][j]] = 1;
    }
  for (int i = 1; i <= n; ++i)
    if (ed[i]) printf("%d ", i);
  printf("\n");
  return 0;
}
