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 N = 2e5 + 5;
const long double eps = 1e-32;
int n, ans[N], cntans;
struct node {
  long double x, y;
  vector<int> id;
  node(long double rx = 0, long double ry = 0) {
    x = rx;
    y = ry;
  }
  bool operator<(const node& rhs) const {
    return x != rhs.x ? x < rhs.x : y < rhs.y;
  }
  node operator-(const node& rhs) const { return node(x - rhs.x, y - rhs.y); }
  long double operator^(const node& rhs) const { return x * rhs.y - y * rhs.x; }
} nd[N];
node stdw[N];
long double cross(node u, node v, node w) { return (v - u) ^ (w - v); }
int tpdw;
long double mn;
bool dwempty() { return tpdw < 2; }
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i <= n; i++) {
    scanf("%d%d", &u, &v);
    nd[i] = node(1e5 / u, 1e5 / v);
    nd[i].id.push_back(i);
  }
  sort(nd + 1, nd + n + 1);
  stdw[++tpdw] = nd[1];
  mn = nd[1].y;
  for (int i = 2; i <= n; i++) {
    if (fabs(nd[i].x - stdw[tpdw].x) <= eps &&
        fabs(nd[i].y - stdw[tpdw].y) <= eps) {
      stdw[tpdw].id.push_back(nd[i].id[0]);
      continue;
    }
    if (nd[i].y >= mn) {
      continue;
    }
    mn = nd[i].y;
    while (!dwempty() && cross(stdw[tpdw - 1], stdw[tpdw], nd[i]) <= eps) {
      --tpdw;
    }
    stdw[++tpdw] = nd[i];
  }
  for (int i = 1; i <= tpdw; i++) {
    for (int j = 0; j < (int)stdw[i].id.size(); j++) {
      ans[++cntans] = stdw[i].id[j];
    }
  }
  sort(ans + 1, ans + cntans + 1);
  for (int i = 1; i <= cntans; i++) {
    printf("%d%c", ans[i], " \n"[i == cntans]);
  }
  return 0;
}
