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;
const int MAX = 200100;
const double eps = 1e-10;
int N;
struct Pt {
  double x, y;
  int i;
};
bool operator<(const Pt &A, const Pt &B) {
  return A.x != B.x ? A.x < B.x : A.y < B.y;
}
Pt pts[MAX];
bool ans[MAX];
double ccw(const Pt &A, const Pt &B, const Pt &C) {
  double x1 = B.x - A.x, y1 = B.y - A.y;
  double x2 = C.x - A.x, y2 = C.y - A.y;
  double w = x1 * y2 - x2 * y1;
  if (w < -eps) return -1;
  if (w > +eps) return +1;
  return 0;
}
int main() {
  scanf("%d", &N);
  for (int i = (0); i < (N); ++i) {
    int s, r;
    scanf("%d%d", &s, &r);
    pts[i] = {1.0 / s, 1.0 / r, i};
  }
  sort(pts, pts + N);
  deque<Pt> D;
  for (int i = (0); i < (N); ++i) {
    while (D.size() >= 2 && ccw(D[D.size() - 2], D[D.size() - 1], pts[i]) < 0)
      D.pop_back();
    D.push_back(pts[i]);
  }
  while (D.size() >= 2 && D[D.size() - 2].y <= D[D.size() - 1].y) D.pop_back();
  for (Pt P : D) ans[P.i] = true;
  int f = 0;
  for (int i = (0); i < (N); ++i)
    if (ans[i]) {
      if (++f > 1) printf(" ");
      printf("%d", i + 1);
    }
  puts("");
  return 0;
}
