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>
int p[200010][3];
int n;
int anst[200010] = {
    0,
};
int ansc = 0;
int comp(const void *pa, const void *pb) {
  int *a = (int *)pa;
  int *b = (int *)pb;
  if (a[1] != b[1]) return b[1] - a[1];
  if (a[2] != b[2]) return b[2] - a[2];
  return a[0] - b[0];
}
long long gcd(long long x, long long y) {
  for (;;) {
    long long t = x % y;
    if (t == 0) break;
    x = y;
    y = t;
  }
  return y;
}
int main() {
  scanf("%d", &n);
  int i;
  for (i = 0; i < n; i++) {
    p[i][0] = i + 1;
    scanf("%d%d", &p[i][1], &p[i][2]);
  }
  qsort(p, n, sizeof(p[0]), comp);
  for (i = 0; i < n; i++) {
    if (ansc == 0) {
      anst[ansc] = i;
      ansc++;
      continue;
    }
    int pr, ppr;
    pr = anst[ansc - 1];
    if (p[pr][1] == p[i][1]) {
      if (p[pr][2] > p[i][2]) continue;
      anst[ansc] = i;
      ansc++;
      continue;
    } else {
      if (p[pr][2] >= p[i][2]) continue;
    }
    while (ansc > 1) {
      pr = anst[ansc - 1];
      ppr = anst[ansc - 2];
      double x00 = (double)p[ppr][2] * p[i][2] * (p[ppr][1] - p[i][1]);
      double x01 = (double)p[ppr][1] * p[i][1] * (p[i][2] - p[ppr][2]);
      double x10 = (long long)p[pr][2] * p[i][2] * (p[pr][1] - p[i][1]);
      double x11 = (long long)p[pr][1] * p[i][1] * (p[i][2] - p[pr][2]);
      if (x00 / x01 > x10 / x11)
        ansc--;
      else
        break;
    }
    anst[ansc] = i;
    ansc++;
  }
  for (i = 0; i < ansc; i++) anst[i] = p[anst[i]][0];
  std::sort(anst, anst + ansc);
  for (i = 0; i < ansc; i++) printf("%d ", anst[i]);
  return 0;
}
