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 INF = 0x3f3f3f3f;
const double eps = 1E-25;
int T;
struct cpoint {
  double x, y;
  int idx;
  cpoint(double a = 0.0, double b = 0.0) { x = a, y = b; }
  cpoint operator-(const cpoint &u) const { return cpoint(x - u.x, y - u.y); }
  double operator*(const cpoint &u) const { return x * u.y - y * u.x; }
  double operator^(const cpoint &u) const { return x * u.x + y * u.y; }
};
int dcmp(double x) {
  if (x < -eps)
    return -1;
  else
    return x > eps;
}
double dis(cpoint p, cpoint q) {
  return sqrt((p.x - q.x) * (p.x - q.x) + (p.y - q.y) * (p.y - q.y));
}
double cross(cpoint o, cpoint p, cpoint q) { return (p - o) * (q - o); }
bool cmp(const cpoint &a, const cpoint &b) {
  return (dcmp(a.y - b.y) == -1) ||
         (dcmp(a.y - b.y) == 0 && dcmp(a.x - b.x) == -1);
}
int graham(int n, cpoint p[], cpoint res[]) {
  if (n == 0) return 0;
  res[0] = p[0];
  if (n == 1) return 1;
  sort(p, p + n, cmp);
  res[1] = p[1];
  if (n == 2) return 2;
  int top = 1, len;
  for (int i = 2; i < n; i++) {
    while (top && dcmp(cross(res[top - 1], res[top], p[i])) < 0) top--;
    res[++top] = p[i];
  }
  len = top;
  res[++top] = p[n - 2];
  for (int i = n - 3; i >= 0; i--) {
    while (top != len && dcmp(cross(res[top - 1], res[top], p[i])) < 0) top--;
    res[++top] = p[i];
  }
  return top;
}
int n, s, r;
cpoint in[200010], out[200010];
vector<int> ans;
int main() {
  int cas = 1;
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      scanf("%d%d", &s, &r);
      in[i].x = 1.0 / s;
      in[i].y = 1.0 / r;
      in[i].idx = i + 1;
    }
    int cnt = graham(n, in, out);
    int cur = 0;
    ans.clear();
    for (int i = 0; i < cnt; i++) {
      ans.push_back(out[cur].idx);
      int pre = (cur - 1 + cnt) % cnt;
      if (dcmp(out[pre].x - out[cur].x) == 1 ||
          (dcmp(out[pre].x - out[cur].x) == 0 &&
           dcmp(out[pre].y - out[cur].y) == 1))
        break;
      cur = pre;
    }
    sort(ans.begin(), ans.end());
    for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}
