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;
int n;
pair<int, int> a[200007];
map<pair<int, int>, int> good;
map<pair<int, int>, int> used;
struct tuhla {
  long long first, second;
  int id;
};
struct tuhla hull[200007];
int nxt[200007];
int prv[200007];
bool f(struct tuhla p1, struct tuhla p2) {
  if (p1.first != p2.first) {
    return (p1.first > p2.first);
  }
  return (p1.second > p2.second);
}
int det(struct tuhla p1, struct tuhla p2, struct tuhla p3) {
  long long k = (p1.second * p2.first * p3.first * p3.second +
                 p1.first * p2.first * p2.second * p3.second +
                 p1.first * p1.second * p2.second * p3.first);
  k -= (p3.second * p2.first * p1.first * p1.second +
        p3.first * p2.first * p2.second * p1.second +
        p3.first * p3.second * p2.second * p1.first);
  if (k > 0) {
    return 1;
  }
  if (k < 0) {
    return -1;
  }
  return 0;
}
void input();
void solve();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}
void input() {
  scanf("%d", &n);
  int i;
  for (i = 0; i < n; i++) {
    scanf("%d%d", &a[i].first, &a[i].second);
    if (used[make_pair(a[i].first, a[i].second)] == 1) {
      continue;
    }
    hull[i].first = a[i].first;
    hull[i].second = a[i].second;
    hull[i].id = i + 1;
    used[make_pair(a[i].first, a[i].second)] = 1;
  }
}
void solve() {
  sort(hull, hull + n, f);
  int i;
  if (det(hull[0], hull[1], hull[2]) < 0) {
    nxt[0] = 2;
    nxt[2] = 1;
    nxt[1] = 0;
    prv[0] = 1;
    prv[1] = 2;
    prv[2] = 0;
  } else {
    nxt[0] = 1;
    nxt[1] = 2;
    nxt[2] = 0;
    prv[0] = 2;
    prv[2] = 1;
    prv[1] = 0;
  }
  int id1, id2;
  for (i = 3; i < n; i++) {
    id1 = i - 1;
    id2 = i - 1;
    while (det(hull[i], hull[id1], hull[nxt[id1]]) < 0) {
      id1 = nxt[id1];
    }
    while (det(hull[i], hull[id2], hull[prv[id2]]) > 0) {
      id2 = prv[id2];
    }
    nxt[i] = id1;
    prv[i] = id2;
    nxt[id2] = i;
    prv[id1] = i;
  }
  vector<int> ans;
  ans.clear();
  int h = 0;
  good[make_pair(hull[0].first, hull[0].second)] = 1;
  while (1) {
    h = nxt[h];
    if (hull[h].second < hull[prv[h]].second) {
      break;
    }
    if (hull[h].second == hull[prv[h]].second &&
        hull[h].first != hull[prv[h]].first) {
      break;
    }
    if (h == 0) {
      break;
    }
    good[make_pair(hull[h].first, hull[h].second)] = 1;
  }
  for (i = 0; i < n; i++) {
    if (good[make_pair(a[i].first, a[i].second)] == 1) {
      ans.push_back(i + 1);
    }
  }
  sort(ans.begin(), ans.end());
  int sz = ans.size();
  for (i = 0; i < sz; i++) {
    printf("%d", ans[i]);
    if (i == sz - 1) {
      printf("\n");
    } else {
      printf(" ");
    }
  }
}
