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;
vector<string> split(const string &s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.push_back(x);
  return v;
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << " = " << a << '\n';
  err(++it, args...);
}
const int MOD = 1000000007;
template <class T>
inline T tmin(T a, T b) {
  return (a < b) ? a : b;
}
template <class T>
inline T tmax(T a, T b) {
  return (a > b) ? a : b;
}
template <class T>
inline void amax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline void amin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline T tabs(T a) {
  return (a > 0) ? a : -a;
}
template <class T>
T gcd(T a, T b) {
  while (b != 0) {
    T c = a;
    a = b;
    b = c % b;
  }
  return a;
}
const int N = 10005;
int n, bit[N];
struct COM {
  int s, r, id;
};
bool comp(COM c1, COM c2) {
  if (c1.s != c2.s) return c1.s < c2.s;
  return c1.r < c2.r;
}
void add(int ind) {
  for (int i = ind; i < N; i += i & -i) {
    bit[i] += 1;
  }
}
int get(int ind) {
  int ans = 0;
  for (int i = ind; i > 0; i -= i & -i) {
    ans += bit[i];
  }
  return ans;
}
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(false);
  int s, r;
  cin >> n;
  vector<COM> coms(n);
  for (int i = (0); i < (n); ++i) {
    cin >> s >> r;
    coms[i].s = N - s;
    coms[i].r = N - r;
    coms[i].id = i + 1;
  }
  sort(coms.begin(), coms.end(), comp);
  vector<int> ans;
  for (int i = (0); i < (n); ++i) {
    if (!get(coms[i].r - 1)) ans.push_back(coms[i].id);
    add(coms[i].r);
  }
  sort(ans.begin(), ans.end());
  int tot = ans.size();
  for (int i = (0); i < (tot); ++i)
    printf("%d%c", ans[i], (i == tot - 1) ? '\n' : ' ');
  return 0;
}
