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 long long INF = 0x3f3f3f3f;
void Open() {}
const long long N = 300010;
struct point {
  long long x, y, id;
  bool operator<(const point& o) const {
    return x > o.x || (x == o.x && y > o.y);
  }
} p[N], ch[N];
int ans[N];
bool cross(point a, point b, point c) {
  return (long long)(a.x - b.x) * (a.y - c.y) * c.x * b.y -
             (long long)(a.y - b.y) * (a.x - c.x) * b.x * c.y <
         0;
}
map<pair<long long, long long>, bool> mp;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    p[i] = (point){x, y, i + 1};
  }
  sort(p, p + n);
  int m = 0;
  for (int i = 0; i < n; i++) {
    if (m >= 1 && ch[m - 1].x == p[i].x) continue;
    while (m > 1 && cross(ch[m - 2], ch[m - 1], p[i])) m--;
    ch[m++] = p[i];
  }
  int xid = -1, yid = -1;
  for (int i = 0; i < m; i++) {
    if (xid == -1 ||
        (ch[i].x > ch[xid].x || (ch[i].x == ch[xid].x && ch[i].y > ch[xid].y)))
      xid = i;
    if (yid == -1 ||
        (ch[i].y > ch[yid].y || (ch[i].y == ch[yid].y && ch[i].x > ch[yid].x)))
      yid = i;
  }
  if (xid > yid) swap(xid, yid);
  int t = 0;
  for (int i = xid; i <= yid; i++) {
    mp[pair<long long, long long>(ch[i].x, ch[i].y)] = 1;
  }
  for (int i = 0; i < n; i++) {
    if (mp[pair<long long, long long>(p[i].x, p[i].y)]) ans[t++] = p[i].id;
  }
  sort(ans, ans + t);
  for (int i = 0; i < t; i++) printf("%d%c", ans[i], " \n"[i == t - 1]);
  return 0;
}
