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;
struct po {
  long double x, y;
  void read() {
    double x0, y0;
    scanf("%lf%lf", &x0, &y0);
    x = 1. / x0, y = 1. / y0;
  }
  inline int operator<(const po &A) const {
    if (x == A.x) return y > A.y;
    return x < A.x;
  }
  po operator-(const po &A) const { return (po){x - A.x, y - A.y}; }
  long double operator*(const po &A) const { return x * A.y - y * A.x; }
  bool operator==(const po &A) const { return (A.x == x) && (A.y == y); }
  long double getang() { return y / x; }
} a[200005], stk[200005];
int n;
map<long double, int> app;
map<po, vector<int> > biao;
vector<int> ans;
int main() {
  scanf("%d", &n);
  for (int i = (int)1; i <= (int)n; i++) {
    a[i].read();
    biao[a[i]].push_back(i);
  }
  sort(a + 1, a + n + 1);
  int top = 0;
  for (int i = (int)1; i <= (int)n; i++) {
    if (a[i] == a[i + 1]) continue;
    if (a[i].x == a[i + 1].x) continue;
    if (app[a[i].y]) continue;
    app[a[i].y] = 1;
    while (top > 1) {
      long double K1 = (a[i] - stk[top - 1]).getang();
      long double K2 = (a[i] - stk[top]).getang();
      if (K2 > 1e-6 || K1 > 1e-6) break;
      if (K2 - K1 >= -1e-12) break;
      top--;
    }
    if (top) {
      long double U = (a[i] - stk[top]).getang();
      if (U > 1e-6) continue;
    }
    stk[++top] = a[i];
  }
  vector<int> A;
  for (int i = (int)1; i <= (int)top; i++) {
    A = biao[stk[i]];
    for (int j = 0; j < (int)A.size(); j++) {
      ans.push_back(A[j]);
    }
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < (int)ans.size(); i++) printf("%d ", ans[i]);
  return 0;
}
