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;
int n;
vector<pair<int, int> > g[10004];
pair<pair<int, int>, int> p[200005];
pair<long double, long double> fin[200005];
vector<int> ind[200005];
bool elem[200005];
const long double eps = 1e-6;
bool lessThanEqual(long double a, long double b) {
  if (a < b + eps)
    return true;
  else
    return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int s, r;
    cin >> s >> r;
    g[s].push_back(make_pair(r, i));
  }
  n = 0;
  for (int i = 1; i <= 10000; ++i) {
    if (g[i].empty()) continue;
    sort(g[i].begin(), g[i].end());
    reverse(g[i].begin(), g[i].end());
    int j = 0;
    while (j < g[i].size() && g[i][j].first == g[i][0].first) {
      p[++n] = make_pair(make_pair(i, g[i][j].first), g[i][j].second);
      j++;
    }
    g[i].clear();
  }
  for (int i = 1; i <= n; ++i)
    g[p[i].first.second].push_back(make_pair(p[i].first.first, p[i].second));
  n = 0;
  for (int i = 1; i <= 10000; ++i) {
    if (g[i].empty()) continue;
    sort(g[i].begin(), g[i].end());
    reverse(g[i].begin(), g[i].end());
    int j = 0;
    while (j < g[i].size() && g[i][j].first == g[i][0].first) {
      p[++n] = make_pair(make_pair(g[i][j].first, i), g[i][j].second);
      j++;
    }
    g[i].clear();
  }
  sort(p + 1, p + n + 1);
  int mxr = 0;
  for (int i = n; i >= 1; i--) {
    int rval = p[i].first.second;
    if (rval < mxr) elem[i] = 1;
    mxr = max(mxr, rval);
  }
  int finct = 0;
  for (int i = 1; i <= n; ++i) {
    if (elem[i]) continue;
    fin[++finct] = make_pair(1.0 / (long double)p[i].first.first,
                             1.0 / (long double)p[i].first.second);
    int j = i;
    while (j <= n && p[j].first == p[i].first) {
      ind[finct].push_back(p[j].second);
      j++;
    }
    i = j - 1;
  }
  vector<int> ans;
  for (int v : ind[1]) ans.push_back(v);
  if (finct > 1)
    for (int v : ind[finct]) ans.push_back(v);
  for (int i = 2; i <= finct - 1; ++i) {
    long double mxr = 0.0;
    for (int j = i + 1; j <= finct; ++j) {
      mxr = max(
          mxr, (fin[i].first - fin[j].first) / (fin[j].second - fin[i].second));
    }
    long double mir = 1e18 + 5;
    for (int j = 1; j <= i - 1; ++j) {
      mir = min(
          mir, (fin[j].first - fin[i].first) / (fin[i].second - fin[j].second));
    }
    if (lessThanEqual(mxr, mir)) {
      for (int v : ind[i]) ans.push_back(v);
    }
  }
  sort(ans.begin(), ans.end());
  for (int v : ans) cout << v << " ";
  cout << '\n';
}
