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 = 200005;
struct node {
  int x, y;
  bool operator<(const node &a) const {
    if (x == a.x) return y < a.y;
    return x < a.x;
  }
} p[MAXN], stk[MAXN];
int call(node a, node b, node c) {
  return (long long)(b.y - a.y) * (c.x - b.x) * c.y * a.x >
         (long long)(c.y - b.y) * (b.x - a.x) * a.y * c.x;
}
set<pair<int, int> > st;
pair<int, int> b[MAXN];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &p[i].x, &p[i].y);
    b[i] = make_pair(p[i].x, p[i].y);
  }
  sort(p + 1, p + n + 1);
  int sz = 0;
  for (int i = 1; i <= n; i++) {
    if (p[i].x != p[i + 1].x) stk[++sz] = p[i];
  }
  memcpy(p, stk, sizeof(stk));
  int top = 0;
  for (int i = 1; i <= sz; i++) {
    while (top > 1 && call(p[i], stk[top - 1], stk[top - 2])) {
      top--;
    }
    if (top == 0 || top > 1)
      stk[top++] = p[i];
    else {
      if (p[i].y >= stk[0].y)
        stk[0] = p[i];
      else
        stk[top++] = p[i];
    }
  }
  for (int i = 0; i < top; i++) st.insert(make_pair(stk[i].x, stk[i].y));
  for (int i = 1; i <= n; i++)
    if (st.count(b[i])) printf("%d ", i);
  printf("\n");
  return 0;
}
