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;
const int Size = 2e5 + 5;
struct Point {
  int x, y, id;
  bool operator<(const Point& p2) const {
    return make_pair(x, y) > make_pair(p2.x, p2.y);
  }
} p[Size], s[Size];
set<pair<int, int> > st;
pair<int, int> b[Size];
int cross(const Point& A, const Point& B, const Point& C) {
  return 1LL * B.y * C.x * (A.x - B.x) * (A.y - C.y) >
         1LL * (A.y - B.y) * (A.x - C.x) * B.x * C.y;
}
int main() {
  int n, m;
  scanf("%d", &n);
  m = n;
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &p[i].x, &p[i].y);
    p[i].id = i + 1;
    b[i] = make_pair(p[i].x, p[i].y);
  }
  sort(p, p + n);
  int top = 0;
  for (int i = 0; i < n; ++i) {
    if (!top || s[top - 1].y < p[i].y) s[top++] = p[i];
  }
  memcpy(p, s, sizeof(s));
  n = top;
  top = 0;
  for (int i = 0; i < n; ++i) {
    while (top >= 2 && cross(s[top - 1], s[top - 2], p[i])) --top;
    s[top++] = p[i];
  }
  for (int i = 0; i < top; ++i) {
    st.insert(make_pair(s[i].x, s[i].y));
  }
  for (int i = 0; i < m; ++i) {
    if (st.count(b[i])) printf("%d ", i + 1);
  }
  return 0;
}
