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 
#include <bits/stdc++.h>
using namespace std;
double eps = 1e-10;
int auto_cnt;
struct node {
  double x, y;
  int index;
  node(double _x = 0, double _y = 0) : x(_x), y(_y) {}
  ~node() {}
  node operator+(const node rhs) const { return node(x + rhs.x, y + rhs.y); }
  node operator-(const node rhs) const { return node(x - rhs.x, y - rhs.y); }
  double operator^(const node rhs) const { return x * rhs.y - y * rhs.x; }
} com[200005];
vector<node> vec;
bool cmpy(node a, node b) {
  if (a.y == b.y) return a.x > b.x;
  return a.y > b.y;
}
bool cmpx(node a, node b) {
  if (a.x == b.x) return a.y > b.y;
  return a.x > b.x;
}
bool cmp(node a, node b) {
  if (a.x == b.x) return a.y < b.y;
  return a.x < b.x;
}
bool cmpp(node a, node b) { return a.index < b.index; }
double cross(node a, node b, node c) { return (b - a) ^ (c - a); }
int main() {
  int n, can;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lf %lf", &com[i].x, &com[i].y);
    com[i].index = i;
  }
  vector<vector<int> > mmap;
  sort(com + 1, com + n + 1, cmpy);
  mmap.push_back(vector<int>());
  mmap[0].push_back(com[1].index);
  com[1].index = 0;
  auto_cnt = 0;
  can = 0;
  for (int i = 2, j = 1; i <= n; i++) {
    if (com[j].y != com[i].y) {
      com[j + 1] = com[i];
      j++;
      auto_cnt++;
      mmap.push_back(vector<int>());
      mmap[auto_cnt].push_back(com[j].index);
      com[j].index = auto_cnt;
    } else if (com[j].x == com[i].x) {
      mmap[auto_cnt].push_back(com[i].index);
      can++;
    } else {
      can++;
    }
  }
  n -= can;
  sort(com + 1, com + n + 1, cmpx);
  can = 0;
  for (int i = 2, j = 1; i <= n; i++) {
    if (com[j].x != com[i].x) {
      com[j + 1] = com[i];
      j++;
    } else {
      can++;
    }
  }
  n -= can;
  for (int i = 1; i <= n; i++) {
    com[i].x = 10000 / com[i].x;
    com[i].y = 10000 / com[i].y;
  }
  sort(com + 1, com + n + 1, cmp);
  vec.push_back(com[1]);
  int cnt = 1;
  for (int i = 2; i <= n; i++) {
    if (cnt < 2) {
      if (com[i].x < vec[cnt - 1].x || com[i].y < vec[cnt - 1].y) {
        vec.push_back(com[i]);
        cnt++;
      }
    } else {
      while (cnt >= 2 &&
             (cross(vec[cnt - 2], vec[cnt - 1], com[i]) + eps < 0)) {
        vec.pop_back();
        cnt--;
      }
      if (com[i].y < vec[cnt - 1].y) {
        vec.push_back(com[i]);
        cnt++;
      }
    }
  }
  sort(vec.begin(), vec.end(), cmpp);
  vector<int> ans;
  for (int i = 0; i < vec.size(); i++) {
    for (int j = 0; j < mmap[vec[i].index].size(); j++) {
      ans.push_back(mmap[vec[i].index][j]);
    }
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) {
    printf("%d%c", ans[i], (i == ans.size() - 1 ? '\n' : ' '));
  }
  return 0;
}
