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 
#include <bits/stdc++.h>
using namespace std;
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch - '0' < 0 || ch - '0' > 9) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch - '0' >= 0 && ch - '0' <= 9) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m;
struct P {
  double x, y;
  int id;
} p[200010], a[200010];
vector<int> v[200010], ans;
bool cmp1(P a, P b) {
  if (a.x != b.x) return a.x < b.x;
  return a.y < b.y;
}
bool dcmp(double x) { return fabs(x) < 1e-9; }
int s[200010], top;
double slope(P a, P b) { return (a.y - b.y) / (a.x - b.x); }
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int x = read(), y = read();
    p[i].x = 1.0 / (double)x;
    p[i].y = 1.0 / (double)y;
    p[i].id = i;
  }
  sort(p + 1, p + n + 1, cmp1);
  a[1] = p[1];
  m = 1;
  v[1].push_back(p[1].id);
  for (int i = 2; i <= n; i++) {
    if (fabs(p[i].x - a[m].x) > 1e-9) a[++m] = p[i];
    if (dcmp(p[i].x - a[m].x) && dcmp(p[i].y - a[m].y)) v[m].push_back(p[i].id);
  }
  s[++top] = 1;
  for (int i = 2; i <= m; i++) {
    while (top > 1 &&
           slope(a[i], a[s[top - 1]]) < slope(a[s[top]], a[s[top - 1]]))
      top--;
    s[++top] = i;
  }
  for (int j = 0; j < v[s[1]].size(); j++) ans.push_back(v[s[1]][j]);
  for (int i = 2; i <= top; i++) {
    if (slope(a[s[i]], a[s[i - 1]]) < 0) {
      for (int j = 0; j < v[s[i]].size(); j++) ans.push_back(v[s[i]][j]);
    } else
      break;
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}
