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 MaxN = 400005;
const double eps = 1e-10;
const double PI = acos(-1.0);
inline int sgn(double x) {
  if (fabs(x) < eps) return 0;
  if (x < 0) return -1;
  return 1;
}
struct Point {
  double x, y;
  int id;
  Point() {}
  Point(double _x, double _y, int c = 0) : x(_x), y(_y) { id = c; }
  Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
  double operator^(const Point &b) const { return x * b.y - y * b.x; }
  double operator*(const Point &b) const { return x * b.x + y * b.y; }
  void trangXY(double B) {
    double tx = x, ty = y;
    x = tx * cos(B) - ty * sin(B);
    y = tx * sin(B) + ty * cos(B);
  }
};
struct Line {
  Point s, e;
  Line() {}
  Line(Point _s, Point _e) : s(_s), e(_e) {}
  pair<int, Point> operator&(const Line &b) const {
    Point ret = s;
    if (sgn((s - e) ^ (b.s - b.e)) == 0)
      if (sgn((s - b.e) ^ (b.s - b.e)) == 0)
        return make_pair(0, ret);
      else
        return make_pair(1, ret);
    double t = ((s - b.s) ^ (b.s - b.e)) / ((s - e) ^ (b.s - b.e));
    ret.x += (e.x - s.x) * t;
    ret.y += (e.y - s.y) * t;
    return make_pair(2, ret);
  }
};
Point list[MaxN];
double dist(const Point &a, const Point &b) { return sqrt((a - b) * (a - b)); }
bool _cmp(const Point &P1, const Point &P2) {
  double temp = (P1 - list[0]) ^ (P2 - list[0]);
  if (sgn(temp) > 0)
    return 1;
  else if (sgn(temp) == 0 && sgn(dist(P1, list[0]) - dist(P2, list[0])) <= 0)
    return 1;
  else
    return 0;
}
void Graham(int Stack[], int &top, int n) {
  Point P0 = list[0];
  int k = 0;
  for (int i = 1; i < n; ++i)
    if (sgn(P0.y - list[i].y) > 0 ||
        (sgn(P0.y - list[i].y) == 0 && sgn(P0.x - list[i].x) > 0)) {
      P0 = list[i];
      k = i;
    }
  swap(list[k], list[0]);
  sort(list + 1, list + n, _cmp);
  if (n == 1) {
    top = 1;
    Stack[0] = 0;
    return;
  }
  if (n == 2) {
    top = 2;
    Stack[0] = 0;
    Stack[1] = 1;
  }
  Stack[0] = 0;
  Stack[1] = 1;
  top = 2;
  for (int i = 2; i < n; ++i) {
    while (top > 1 && sgn((list[Stack[top - 1]] - list[Stack[top - 2]]) ^
                          (list[i] - list[Stack[top - 2]])) <= 0)
      --top;
    Stack[top++] = i;
  }
}
int ans[MaxN], ansnum;
int zans[MaxN], cou;
int main() {
  int n;
  int s, r;
  double minn = 100000000.0;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &s, &r);
    list[i].x = 1.0 / s;
    list[i].y = 1.0 / r;
    list[i].id = i + 1;
    minn = min(minn, 1.0 / s);
  }
  Graham(ans, ansnum, n);
  cou = 0;
  zans[cou++] = list[ans[0]].id;
  for (int i = ansnum - 1; i >= 0; --i) {
    zans[cou++] = list[ans[i]].id;
    if (sgn(list[ans[i]].x - minn) == 0) break;
  }
  sort(zans, zans + cou);
  for (int i = 0; i < cou; ++i) printf("%d ", zans[i]);
  puts("");
  return 0;
}
