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 double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const long long mod = (long long)(1e9 + 7);
const int MAX_N = 200010;
const double eps = 1e-7;
int n, m;
int vis[MAX_N];
inline int sgn(double x) {
  if (fabs(x) <= eps)
    return 0;
  else if (x > 0)
    return 1;
  else
    return -1;
}
struct Point {
  int tx, ty, id;
  double x, y;
  Point() {}
  Point(double _x, double _y) : x(_x), y(_y) {}
  Point operator-(const Point& rhs) const {
    return Point(x - rhs.x, y - rhs.y);
  }
  double cross(const Point& rhs) const { return x * rhs.y - y * rhs.x; }
  bool operator<(const Point& rhs) const {
    if (!sgn(x - rhs.x)) return y < rhs.y;
    return x < rhs.x;
  }
} P[MAX_N], V[MAX_N], leftCorner;
int andrew() {
  sort(P, P + n);
  int k = 0;
  for (int i = 0; i < n; ++i) {
    if (i && !sgn(P[i].x - P[i - 1].x)) continue;
    while (k > 1 && sgn((V[k - 1] - V[k - 2]).cross(P[i] - V[k - 2])) < 0) k--;
    V[k++] = P[i];
  }
  return k;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    P[i].tx = x, P[i].ty = y, P[i].id = i + 1;
    P[i].x = 100000.0 / x, P[i].y = 100000.0 / y;
  }
  memset(vis, 0, sizeof(vis));
  m = andrew();
  for (int i = 0; i < m; ++i) {
    if (!i || sgn(V[i].y - V[i - 1].y) < 0) vis[V[i].id] = 1;
  }
  for (int i = 0; i < n; ++i) {
    if (i && !sgn(P[i].x - P[i - 1].x) && !sgn(P[i].y - P[i - 1].y) &&
        vis[P[i - 1].id]) {
      vis[P[i].id] = 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (vis[i]) printf("%d ", i);
  }
  puts("");
  return 0;
}
