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;
int read_int() {
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') {
    c = getchar();
  }
  bool sign = 0;
  if (c == '-') {
    sign = 1;
    c = getchar();
  }
  int ret = 0;
  while (c >= '0' && c <= '9') {
    ret = 10 * ret + c - 48;
    c = getchar();
  }
  if (sign) return -ret;
  return ret;
}
int read_string(char *s) {
  int n = 0;
  char c = getchar();
  while (c < 'a' || c > 'z') c = getchar();
  while (c >= 'a' && c <= 'z') {
    s[n] = c - 'a';
    n++;
    c = getchar();
  }
  return n;
}
const double EPS = 1e-12;
struct PT {
  double x, y;
  PT() {}
  PT(double x, double y) : x(x), y(y) {}
  bool operator<(const PT &rhs) const {
    return make_pair(y, x) < make_pair(rhs.y, rhs.x);
  }
  bool operator==(const PT &rhs) const {
    return make_pair(y, x) == make_pair(rhs.y, rhs.x);
  }
};
double cross(PT p, PT q) { return p.x * q.y - p.y * q.x; }
double area2(PT a, PT b, PT c) {
  return cross(a, b) + cross(b, c) + cross(c, a);
}
void ConvexHull(vector<PT> &pts) {
  sort(pts.begin(), pts.end());
  pts.erase(unique(pts.begin(), pts.end()), pts.end());
  vector<PT> up, dn;
  for (int i = 0; i < pts.size(); i++) {
    while (up.size() > 1 && area2(up[up.size() - 2], up.back(), pts[i]) >= 0)
      up.pop_back();
    while (dn.size() > 1 && area2(dn[dn.size() - 2], dn.back(), pts[i]) <= 0)
      dn.pop_back();
    up.push_back(pts[i]);
    dn.push_back(pts[i]);
  }
  pts = dn;
  for (int i = (int)up.size() - 2; i >= 1; i--) pts.push_back(up[i]);
}
int main() {
  srand(time(NULL));
  int t = 1;
  while (t-- > 0) {
    int n = read_int();
    map<pair<int, int>, vector<int> > a;
    vector<PT> b;
    for (int i = 1; i <= n; i++) {
      int x = read_int();
      int y = read_int();
      b.push_back(PT(1 / (double)x, 1 / (double)y));
      a[make_pair(x, y)].push_back(i);
    }
    ConvexHull(b);
    set<int> c;
    int p = 0;
    for (int i = 1; i < b.size(); i++) {
      if (b[i].x < b[p].x || (b[i].x == b[p].x && b[i].y < b[p].y)) {
        p = i;
      }
    }
    int x1 = (int)(1 / b[0].x + 0.5), y1 = (int)(1 / b[0].y + 0.5);
    for (int j = 0; j < a[make_pair(x1, y1)].size(); j++) {
      c.insert(a[make_pair(x1, y1)][j]);
    }
    for (int i = p; i < b.size(); i++) {
      x1 = (int)(1 / b[i].x + 0.5), y1 = (int)(1 / b[i].y + 0.5);
      for (int j = 0; j < a[make_pair(x1, y1)].size(); j++) {
        c.insert(a[make_pair(x1, y1)][j]);
      }
    }
    for (set<int>::iterator it = c.begin(); it != c.end(); it++) {
      printf("%d ", *it);
    }
    printf("\n");
  }
  return 0;
}
