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 
5
10 50
10 50
10 50
10 50
10 50
 2
10000 10000
10000 10000
 3
10000 10000
10000 10000
10000 10000
 3
1000 3000
1500 1500
3000 1000
 43
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 1
1 1
 2
1 1
1 1
 44
78 52
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 3
1000 3000
1499 1499
3000 1000
 5
50 50
49 50
50 49
49 49
50 1
 18
82 38
33 69
33 69
33 69
33 69
33 69
82 38
33 69
33 69
33 69
82 38
33 69
33 69
82 38
82 38
33 69
33 69
33 69
 1
10000 9999
 3
1000 3000
1501 1501
3000 1000
 2
1 2
2 1

1 2 3 4 5
 1 2
 1 2 3
 1 2 3
 1 3 16 24 29 31 32 37 41 42
 1
 1 2
 1
 1 3
 1
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
 1
 1 2 3
 1 2 

#include <bits/stdc++.h>
using namespace std;
inline long long ma(long long a, long long b) { return ((a - b > 0) ? a : b); }
inline long long mi(long long a, long long b) { return ((a - b > 0) ? b : a); }
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
bool Q;
struct line {
  mutable long long k, m, p;
  bool operator<(const line& o) const { return Q ? p < o.p : k < o.k; }
};
struct LineContainer : multiset<line> {
  const long long inf = LLONG_MAX;
  inline long long div(long long a, long long b) {
    return a / b - ((a ^ b) < 0 && a % b);
  }
  inline bool isect(iterator x, iterator y) {
    if (y == end()) {
      x->p = inf;
      return false;
    }
    if (x->k == y->k)
      x->p = x->m > y->m ? inf : -inf;
    else
      x->p = div(y->m - x->m, x->k - y->k);
    return x->p >= y->p;
  }
  inline void add(long long k, long long m) {
    auto z = insert({k, m, 0}), y = z++, x = y;
    while (isect(y, z)) z = erase(z);
    if (y != begin() && isect(--x, y)) isect(x, erase(y));
    while ((y = x) != begin() && (--x)->p >= y->p) isect(x, erase(y));
  }
  inline pair<long long, long long> query(long long x) {
    assert(!empty());
    Q = 1;
    auto l = *lower_bound({0, 0, x});
    Q = 0;
    return {l.k, l.m};
  }
};
struct T {
  vector<long long> v;
  T() {}
  T(vector<long long> v) : v(v) {}
};
struct ST {
  long long n, ans;
  T t[2 * 200010];
  inline T Op(T& val1, T& val2) {
    vector<long long> v;
    for (int(i) = (0); (i) < (val1.v.size()); (i)++) v.push_back(val1.v[i]);
    for (int(i) = (0); (i) < (val2.v.size()); (i)++) v.push_back(val2.v[i]);
    sort((v).begin(), (v).end());
    T ty;
    ty.v = v;
    return ty;
  }
  inline long long Op1(T& val1, long long& k) {
    ans = 0;
    ans += (upper_bound((val1.v).begin(), (val1.v).end(), k) - val1.v.begin());
    return ans;
  }
  inline void build() {
    for (int(i) = (n - 1); (i) >= (1); (i)--)
      t[i] = Op(t[i << 1], t[i << 1 | 1]);
  }
  inline long long que(long long l, long long r, long long k) {
    long long ans = 0;
    for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
      if (l & 1) ans += Op1(t[l++], k);
      if (r & 1) ans += Op1(t[--r], k);
    }
    return ans;
  }
} st;
long long ar[200010], dp[200010], a[200010], b[200010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  long long n;
  cin >> n;
  ST st1, st2;
  st1.n = n;
  st2.n = n;
  vector<pair<long long, pair<long long, long long> > > v1, v2;
  for (int(i) = (0); (i) < (n); (i)++) {
    long long a, b;
    cin >> a >> b;
    v1.push_back({a, {b, i}});
  }
  sort((v1).begin(), (v1).end());
  for (int(i) = (0); (i) < (n); (i)++) {
    long long a = v1[i].first, b = v1[i].second.first;
    st1.t[i + n].v.push_back(a), st2.t[i + n].v.push_back(b);
  }
  st1.build();
  st2.build();
  vector<long long> froz;
  for (int(i) = (0); (i) < (n - 1); (i)++) {
    long long l = i + 1, r = n;
    long long a = v1[i].first, b = v1[i].second.first, id = v1[i].second.second,
              ta = r - l;
    long long nxt1 = v1[i + 1].first, nxt2 = v1[i + 1].second.first;
    if (nxt1 == a && i != n - 2 && b < nxt2) continue;
    if (nxt1 == a && b == nxt2) continue;
    long long val1 = st1.que(l, r, a), val2 = st2.que(l, r, b);
    if (val1 == ta || val2 == ta) froz.push_back(id);
  }
  froz.push_back(v1[n - 1].second.second);
  sort((froz).begin(), (froz).end());
  for (auto xd : froz) cout << xd + 1 << " ";
  cout << endl;
  return 0;
}
