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;
template <typename T>
inline bool chkmin(T& x, T y) {
  return x > y ? x = y, 1 : 0;
}
template <typename T>
inline bool chkmax(T& x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <typename T>
inline void read(T& x) {
  char c;
  int sign = 1;
  x = 0;
  do {
    c = getchar();
    if (c == '-') sign = -1;
  } while (!isdigit(c));
  do {
    x = x * 10 + c - '0';
    c = getchar();
  } while (isdigit(c));
  x *= sign;
}
const int N = 3e5 + 50;
struct DATA {
  double x, y;
  int id;
  DATA() {}
  DATA(double _x, double _y) { x = _x, y = _y; }
  DATA operator-(DATA b) const { return DATA(x - b.x, y - b.y); }
} a[N], s[N];
const bool cmp(DATA a, DATA b) { return a.x < b.x || a.x == b.x && a.y < b.y; }
const bool cmp2(DATA a, DATA b) { return a.y < b.y || a.y == b.y && a.x < b.x; }
inline bool check(DATA a, DATA b) { return a.x * b.y - a.y * b.x >= 0; }
int top, n;
bool vis[N];
int main() {
  read(n);
  int m = n;
  for (register int i = 1; i <= n; ++i) read(a[i].x), read(a[i].y), a[i].id = i;
  sort(a + 1, a + 1 + n, cmp2);
  for (register int i = 1; i <= n; ++i) {
    while (top && a[i].x > a[top].x && a[i].y == a[top].y) --top;
    s[++top] = a[i];
  }
  for (register int i = 1; i <= top; ++i) a[i] = s[i];
  n = top;
  top = 0;
  sort(a + 1, a + 1 + n, cmp);
  for (register int i = 1; i <= n; ++i) {
    while (top && a[i].y > a[top].y && a[i].x == a[top].x) --top;
    s[++top] = a[i];
  }
  for (register int i = 1; i <= top; ++i) a[i] = s[i];
  n = top;
  top = 0;
  for (register int i = 1; i <= n; ++i)
    a[i].x = 1.0 / a[i].x, a[i].y = 1.0 / a[i].y;
  sort(a + 1, a + 1 + n, cmp);
  for (register int i = 1; i <= n; ++i) {
    while (top > 1 && check(a[i] - s[top - 1], s[top] - s[top - 1])) --top;
    if (!top || a[i].y < s[top].y || a[i].x == s[top].x && a[i].y == s[top].y)
      s[++top] = a[i];
  }
  for (register int i = 1; i <= top; ++i) vis[s[i].id] = 1;
  for (register int i = 1; i <= m; ++i)
    if (vis[i]) printf("%d ", i);
  return 0;
}
