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 n, m, p, l, ans[200005], ans2[200005], q[200005], hd, tl;
struct d {
  int x, y, id;
} s[200005], t[200005];
bool cmp(d a, d b) { return a.x != b.x ? a.x > b.x : a.y > b.y; }
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); i++) {
    scanf("%d %d", &s[i].x, &s[i].y);
    s[i].id = i + 1;
  }
  sort(s, s + n, cmp);
  for (int i = (0); i < (n); i++) {
    m = s[i].y;
    int x = s[i].x;
    while (i < n && s[i].x == x) {
      if (s[i].y == m && s[i].y > p) t[l++] = s[i];
      i++;
    }
    p = max(p, m);
    i--;
  }
  for (int i = (0); i < (l); i++) {
    int j = i;
    while (i < n && t[i].x == t[j].x && t[i].y == t[j].y) i++;
    i--;
    while (hd + 1 < tl &&
           (long long)t[q[tl - 1]].x * t[q[tl - 2]].y *
                   (t[q[tl - 1]].y - t[i].y) * (t[i].x - t[q[tl - 2]].x) >
               (long long)t[q[tl - 2]].x * t[q[tl - 1]].y *
                   (t[i].x - t[q[tl - 1]].x) * (t[q[tl - 2]].y - t[i].y))
      tl--;
    q[tl++] = i;
  }
  for (int i = (0); i < (tl); i++) ans[q[i]] = 1;
  int ptr = 0;
  for (int i = (0); i < (l); i++) {
    while (ptr < tl && q[ptr] < i) ptr++;
    if (ptr < tl && t[i].x == t[q[ptr]].x && t[i].y == t[q[ptr]].y)
      ans2[t[i].id] = 1;
  }
  for (int i = (1); i <= (n); i++)
    if (ans2[i]) printf("%d\n", i);
  scanf("\n");
}
