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 long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f * x;
}
const long long maxn = 3e5 + 5;
long long n, tot, ca[maxn];
struct node {
  long long a, b, id;
  bool friend operator<(node x, node y) {
    return x.a == y.a ? x.b > y.b : x.a > y.a;
  }
} p[maxn], t[maxn];
long long pos[maxn];
long long sta[maxn], top;
inline double cal(long long x, long long y) {
  return 1.0 * t[x].a * t[y].a * (t[y].b - t[x].b) /
         (1.0 * t[x].b * t[y].b * (t[y].a - t[x].a));
}
signed main() {
  n = read();
  for (long long i = 1; i <= n; ++i)
    p[i].a = read(), p[i].b = read(), p[i].id = i;
  sort(p + 1, p + n + 1);
  long long max_b = p[1].b;
  t[++tot] = p[1];
  pos[1] = 1;
  for (long long i = 2; i <= n; ++i) {
    if (p[i].b <= max_b) continue;
    max_b = max(max_b, p[i].b);
    t[++tot] = p[i];
    pos[tot] = i;
  }
  sta[1] = 1, sta[2] = 2;
  top = 2;
  for (long long i = 3; i <= tot; ++i) {
    while (top > 1 && cal(sta[top], i) < cal(sta[top - 1], sta[top])) --top;
    sta[++top] = i;
  }
  for (long long i = 1; i <= top; ++i) {
    long long nw = sta[i];
    long long ta = t[nw].a, tb = t[nw].b;
    for (long long j = pos[nw]; j <= n; ++j)
      if (p[j].a == ta && p[j].b == tb)
        ca[p[j].id] = 1;
      else
        break;
  }
  for (long long i = 1; i <= n; ++i)
    if (ca[i]) printf("%lld ", i);
  return 0;
}
