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;
struct point {
  int x, y;
  point() {}
  point(int _x, int _y) : 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 (p.x != x) return x > p.x;
    return y > p.y;
  }
  bool operator==(const point &p) const { return x == p.x && y == p.y; }
  int operator*(const point &p) const { return x * p.y - y * p.x; }
};
int n, m, X, cnt, las, top[10005];
point p[200005], q[10005], ch[10005];
bool ans[10005], fi = true;
inline bool judge(point A, point B, point C) {
  point p = C - B, q = B - A;
  return 1LL * A.x * C.y * q.y * p.x > 1LL * A.y * C.x * p.y * q.x;
}
inline int ConvexHull(point p[], int n) {
  int m = 0;
  for (int i = 1; i <= n; i++) {
    while (m > 1 && judge(ch[m - 1], ch[m], p[i])) --m;
    ch[++m] = p[i];
  }
  return m;
}
void out(int x) {
  fi ? fi = false : printf(" ");
  printf("%d", x);
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &p[i].x, &p[i].y);
    X = max(X, p[i].x);
    top[p[i].x] = max(top[p[i].x], p[i].y);
  }
  for (int i = X; i; i--)
    if (top[i] > las) q[++cnt] = point(i, las = top[i]);
  m = ConvexHull(q, cnt);
  for (int i = 1; i <= m; i++) ans[ch[i].x] = true;
  for (int i = 1; i <= n; i++)
    if (ans[p[i].x] && top[p[i].x] == p[i].y) out(i);
  puts("");
  return 0;
}
