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;
const int N = 200005;
const double inf = 1e18;
struct data {
  long long r, s;
  int id;
} dat[N], tmp[N];
long long mx[N];
int idx, n;
bool flag[N], flag2[N];
int p[N];
bool cmp(data a, data b) {
  if (a.r < b.r) return true;
  if (a.r == b.r && a.s < b.s) return true;
  return false;
}
double f(int a, int b) {
  return (double)dat[a].r * (dat[a].s - dat[b].s) /
         (double)(dat[a].s * (dat[b].r - dat[a].r));
}
double f2(int a, int b) {
  return (double)dat[b].r * (dat[a].s - dat[b].s) /
         (double)(dat[b].s * (dat[b].r - dat[a].r));
}
double getL(int ix) {
  int l = 1, r = ix - 1, m1, m2;
  if (l > r) return 0;
  double val1, val2, ret = -inf;
  while (r - l > 5) {
    m1 = (l + r) / 2;
    m2 = m1 + 1;
    val1 = f(m1, ix);
    val2 = f(m2, ix);
    if (val1 >= val2) {
      r = m2;
    } else {
      l = m1;
    }
  }
  for (int i = l; i <= r; i++) {
    ret = max(ret, f(i, ix));
  }
  return ret;
}
double getR(int ix) {
  int l = ix + 1, r = idx - 1, m1, m2;
  if (l > r) return 0;
  double val1, val2, ret = inf;
  while (r - l > 5) {
    m1 = (l + r) / 2;
    m2 = m1 + 1;
    val1 = f2(ix, m1);
    val2 = f2(ix, m2);
    if (val1 >= val2) {
      l = m1;
    } else {
      r = m2;
    }
  }
  for (int i = l; i <= r; i++) {
    ret = min(ret, f2(ix, i));
  }
  return ret;
}
int find(int x) {
  if (p[x] == x) return x;
  return p[x] = find(p[x]);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) p[i] = i;
  for (int i = 1; i <= n; i++) {
    cin >> tmp[i].r >> tmp[i].s;
    tmp[i].id = i;
  }
  sort(tmp + 1, tmp + n + 1, cmp);
  tmp[n + 1].r = tmp[n + 1].s = 100000;
  mx[n + 1] = 0;
  idx = 1;
  for (int i = n; i >= 1; i--) {
    mx[i] = max(mx[i + 1], tmp[i].s);
    if (tmp[i].s == tmp[i + 1].s && tmp[i].r == tmp[i + 1].r) {
      flag[i] = true;
      p[tmp[i].id] = find(p[tmp[i + 1].id]);
      continue;
    }
    if (tmp[i].s <= mx[i + 1]) flag[i] = true;
  }
  for (int i = 1; i <= n; i++) {
    if (flag[i]) continue;
    dat[idx++] = tmp[i];
  }
  vector<int> ans;
  for (int i = 1; i < idx; i++) {
    if (i == 1 || i == idx - 1) {
      flag2[dat[i].id] = true;
    } else if (getL(i) <= getR(i)) {
      flag2[dat[i].id] = true;
    }
  }
  int pa;
  for (int i = 1; i <= n; i++) {
    pa = find(p[tmp[i].id]);
    if (!flag2[pa]) continue;
    ans.push_back(tmp[i].id);
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) {
    if (i > 0) cout << " ";
    cout << ans[i];
  }
  cout << endl;
  return 0;
}
