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;
const int MAXN = 200010;
int s[MAXN], r[MAXN];
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
pair<long long, long long> get(int i, int j) {
  long long tmp1 = (long long)(r[j] - r[i]) * s[i] * s[j];
  long long tmp2 = (long long)r[i] * r[j] * (s[i] - s[j]);
  long long tmp = gcd(tmp1, tmp2);
  tmp1 /= tmp;
  tmp2 /= tmp;
  return make_pair(tmp1, tmp2);
}
int com(pair<long long, long long> t1, pair<long long, long long> t2) {
  int neg1 = 1;
  int neg2 = 1;
  long long t11 = t1.first;
  long long t12 = t1.second;
  long long t21 = t2.first;
  long long t22 = t2.second;
  if ((t11 >= 0) != (t12 >= 0)) neg1 = -1;
  if ((t21 >= 0) != (t22 >= 0)) neg2 = -1;
  if (neg1 < neg2) return -1;
  if (neg1 > neg2) return 1;
  t11 = abs(t11);
  t12 = abs(t12);
  t21 = abs(t21);
  t22 = abs(t22);
  if (neg1 > 0) {
    if (t11 == t21 && t12 == t22)
      return 0;
    else if ((double)t11 / t12 < (double)t21 / t22)
      return -1;
    else if ((double)t11 / t12 > (double)t21 / t22)
      return 1;
    else
      return 0;
  } else {
    if (t11 == t21 && t12 == t22)
      return 0;
    else if ((double)t11 / t12 < (double)t21 / t22)
      return 1;
    else if ((double)t11 / t12 > (double)t21 / t22)
      return -1;
    else
      return 0;
  }
}
int a[MAXN], b[MAXN];
vector<int> vec[10010];
pair<long long, long long> pp1[10010], pp2[10010];
int sta[10010];
int f[10010];
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    for (int i = 1; i <= 10000; i++) vec[i].clear();
    for (int i = 0; i < n; i++) {
      scanf("%d%d", &a[i], &b[i]);
      vec[a[i]].push_back(b[i]);
    }
    int m = 0;
    for (int i = 1; i <= 10000; i++) {
      if (vec[i].size() == 0) continue;
      sort(vec[i].begin(), vec[i].end());
      reverse(vec[i].begin(), vec[i].end());
      s[m] = i;
      r[m++] = vec[i][0];
    }
    for (int i = 0; i < m; i++) pp1[i] = make_pair(0, 1);
    int top = 0;
    sta[top++] = 0;
    for (int i = 1; i < m; i++) {
      while (top > 1 && com(get(i, sta[top - 2]), get(i, sta[top - 1])) >= 0)
        top--;
      pair<long long, long long> ttt = get(i, sta[top - 1]);
      if (com(ttt, pp1[i]) > 0) pp1[i] = ttt;
      sta[top++] = i;
    }
    top = 0;
    sta[top++] = m - 1;
    for (int i = m - 2; i >= 0; i--) {
      while (top > 1 && com(get(i, sta[top - 2]), get(i, sta[top - 1])) <= 0)
        top--;
      pp2[i] = get(i, sta[top - 1]);
      sta[top++] = i;
    }
    memset(f, -1, sizeof(f));
    for (int i = 0; i < m; i++) {
      if (i == m - 1) {
        f[s[i]] = r[i];
      } else {
        if (com(pp1[i], pp2[i]) <= 0) f[s[i]] = r[i];
      }
    }
    vector<int> ans;
    for (int i = 0; i < n; i++)
      if (b[i] == f[a[i]]) ans.push_back(i + 1);
    int sz = ans.size();
    for (int i = 0; i < sz; i++) {
      printf("%d", ans[i]);
      if (i < sz - 1)
        printf(" ");
      else
        printf("\n");
    }
  }
  return 0;
}
