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 = 2e5 + 100;
const double maxv = 1e6;
const double eps = 1e-9;
inline double sgn(double v) {
  if (abs(v) < eps)
    return 0;
  else
    return v < 0 ? -1 : 1;
}
struct Point {
  double x, y;
  int idx;
  Point(double _x = 0, double _y = 0) {
    x = _x;
    y = _y;
  }
  Point operator-(const Point &p) const { return Point(x - p.x, y - p.y); }
  bool operator<(const Point &p) const {
    if (sgn(x - p.x) != 0)
      return x < p.x;
    else
      return y < p.y;
  }
} pt[maxn];
bool winner[maxn];
int stk[maxn], stnum;
double cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
void convex(int n) {
  int i, j;
  sort(pt, pt + n);
  stnum = 0;
  for (i = 0, j = 0; i < n; i++)
    if (i == 0 || pt[i].y >= pt[j].y) pt[j++] = pt[i];
  n = j;
  for (i = 0; i < n; i++) {
    if (stnum > 2 && sgn(cross(pt[stk[stnum - 1]] - pt[stk[stnum - 2]],
                               pt[i] - pt[stk[stnum - 1]])) <= 0)
      stnum--;
    stk[stnum++] = i;
  }
  while (stnum > 2 &&
         sgn(cross(pt[stk[stnum - 1]] - pt[stk[stnum - 2]], Point(1, 0))) <= 0)
    stnum--;
  for (i = 0; i < stnum - 1; i++)
    if (sgn(cross(pt[stk[i + 1]] - pt[stk[i]], Point(1, 0))) != 0) break;
  for (; i < stnum; i++) winner[pt[stk[i]].idx] = true;
}
int main() {
  int n, i;
  int s, r;
  memset(winner, false, sizeof(winner));
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &s, &r);
    pt[i].x = maxv / s;
    pt[i].y = maxv / r;
    pt[i].idx = i + 1;
  }
  convex(n);
  for (i = 1; i <= n; i++)
    if (winner[i]) printf("%d ", i);
  return 0;
}
