Tavas is a cheerleader in the new sports competition named "Pashmaks".

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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;
pair<long double, long double> A[200666];
int st[200666], ind[200666];
long double val[200666];
bool comp(int i1, int i2) { return A[i1] < A[i2]; }
bool eql(long double v1, long double v2) {
  return (v1 - v2 > -((long double)0.00000000000001) &&
          v1 - v2 < ((long double)0.00000000000001));
}
long double bs(long double vmax, pair<long double, long double> v1,
               pair<long double, long double> v2) {
  long double pos, step;
  for (step = 1; step <= vmax; step *= (long double)2.0)
    ;
  for (pos = 0; step > ((long double)0.0000000000000000001);
       step /= (long double)2.0)
    if (pos + step <= vmax) {
      long double val_1 = 1.0 / v1.first + (pos + step) / (v1.second);
      long double val_2 = 1.0 / v2.first + (pos + step) / (v2.second);
      if (val_2 < val_1 || eql(val_2, val_1)) pos += step;
    }
  return pos;
}
int main() {
  int n, i;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    int v1, v2;
    scanf("%d%d", &v1, &v2);
    A[i].first = v1;
    A[i].second = v2;
    ind[i] = i;
  }
  sort(ind + 1, ind + n + 1, comp);
  int k = 1;
  st[1] = ind[1];
  val[1] = ((long double)1000000000000000);
  for (i = 2; i <= n; ++i) {
    int l = ind[i];
    while (true && k > 0) {
      long double v1 = 1.0 / A[st[k]].first + val[k] / (A[st[k]].second);
      long double v2 = 1.0 / A[l].first + val[k] / (A[l].second);
      if (eql(v1, v2)) break;
      if (v2 < v1) {
        --k;
        continue;
      }
      break;
    }
    if (k == 0) {
      st[++k] = l;
      continue;
    }
    st[++k] = l;
    val[k] = bs(val[k - 1], A[st[k - 1]], A[l]);
  }
  sort(st + 1, st + k + 1);
  for (i = 1; i <= k; ++i) cout << st[i] << " ";
  return 0;
}
