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 eps = 1e-22;
struct node {
  double x, y;
  int w;
} a[200010], b[200010];
int f[200010], ans[200010], n, top = 0;
inline double dcmp(double x) {
  if (fabs(x) < eps)
    return 0;
  else if (x > 0)
    return 1;
  else if (x < 0)
    return -1;
}
bool cmp(node a, node b) {
  if (dcmp(a.x - b.x) == -1)
    return true;
  else if (dcmp(a.x - b.x) == 1)
    return false;
  else if (dcmp(a.y - b.y) >= 0)
    return true;
  else
    return false;
}
double cross(node x1, node x2, node y1, node y2) {
  double dx1 = x1.x - x2.x, dy1 = x1.y - x2.y, dx2 = y1.x - y2.x,
         dy2 = y1.y - y2.y;
  return dx1 * dy2 - dx2 * dy1;
}
bool same(node a, node b) {
  if (dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0) return true;
  return false;
}
int main() {
  scanf("%d", &n);
  int jx, jy;
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &jx, &jy);
    a[i].x = (double)(1.0 / jx);
    a[i].y = (double)(1.0 / jy);
    a[i].w = i;
    b[i] = a[i];
  }
  a[0].w = 0;
  a[0].y = 1e9;
  a[0].x = 0;
  a[n + 1].w = n + 1;
  a[n + 1].x = 1e9;
  a[n + 1].y = 0;
  b[0] = a[0];
  b[n + 1] = a[n + 1];
  sort(a + 1, a + 1 + n, cmp);
  memset(f, 0, sizeof f);
  top = 0;
  for (int i = 0; i <= n + 1; i++) {
    while (top > 1 && dcmp(cross(b[a[f[top]].w], b[a[f[top - 1]].w], a[i],
                                 b[a[f[top - 1]].w])) <= 0)
      top--;
    top++;
    f[top] = i;
  }
  for (int i = 1; i <= top; i++) f[i] = a[f[i]].w;
  sort(f + 1, f + 1 + top);
  for (int i = 1; i <= top; i++)
    if (f[i] != 0 && f[i] != n + 1) printf("%d ", f[i]);
}
