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;
inline int read() {
  int a = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    a = (a << 3) + (a << 1) + (c ^ '0');
    c = getchar();
  }
  return a;
}
const int MAXN = 300010;
struct point {
  int a, b, ind;
} now[MAXN];
int nxt[MAXN], pre[MAXN], S[MAXN], tl;
bool can[MAXN];
bool cmp(point a, point b) {
  if (a.a == b.a) return a.b > b.b;
  return a.a > b.a;
}
long double calcK(point a, point b) {
  return (long double)a.a * b.a * (b.b - a.b) / a.b / b.b / (b.a - a.a);
}
int main() {
  int N = read();
  for (int i = 1; i <= N; i++) {
    now[i].a = read();
    now[i].b = read();
    now[i].ind = i;
    nxt[i] = i + 1;
    pre[i] = i - 1;
  }
  sort(now + 1, now + N + 1, cmp);
  int maxB = now[1].b;
  for (int i = 2; i <= N; i++)
    if (now[i].b <= maxB) {
      nxt[pre[i]] = nxt[i];
      pre[nxt[i]] = pre[i];
    } else
      maxB = now[i].b;
  S[0] = 1;
  tl = 1;
  for (int i = nxt[1]; i <= N; i = nxt[i]) {
    while (tl != 1 && calcK(now[i], now[S[tl - 1]]) <
                          calcK(now[S[tl - 2]], now[S[tl - 1]]))
      tl--;
    S[tl++] = i;
  }
  for (int i = 0; i < tl; i++) {
    can[now[S[i]].ind] = 1;
    for (int j = S[i] + 1;
         j <= N && now[S[i]].a == now[j].a && now[S[i]].b == now[j].b; j++)
      can[now[j].ind] = 1;
  }
  for (int i = 1; i <= N; i++)
    if (can[i]) printf("%d ", i);
  return 0;
}
