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>
void gn(int &x) {
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') x = x * 10 + c - '0';
}
int qp(int a, long long b) {
  int ans = 1;
  do {
    if (b & 1) ans = 1ll * ans * a % 1000000007;
    a = 1ll * a * a % 1000000007;
  } while (b >>= 1);
  return ans;
}
using namespace std;
int n;
struct ren {
  int id, k, b;
} a[222222];
int cmp(const ren &x, const ren &y) {
  if (x.k == y.k) return x.b > y.b;
  return x.k < y.k;
}
int ok[222222] = {0};
int stk[222222], top;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", &a[i].k, &a[i].b), a[i].id = i;
  sort(a + 1, a + 1 + n, cmp);
  top = 0;
  for (int i = 1; i <= n; i++) {
    if (i > 1 && a[i].k == a[i - 1].k) continue;
    while (
        top >= 2 &&
        1ll *
                ((a[stk[top - 1]].b) * (a[stk[top]].b) -
                 (a[stk[top - 1]].b) * (a[i].b)) *
                ((a[stk[top]].k) * (a[i].k) - (a[stk[top - 1]].k) * (a[i].k)) <
            1ll *
                ((a[stk[top - 1]].b) * (a[i].b) - (a[stk[top]].b) * (a[i].b)) *
                ((a[stk[top - 1]].k) * (a[i].k) -
                 (a[stk[top - 1]].k) * (a[stk[top]].k)))
      top--;
    stk[++top] = i;
  }
  for (int i = 1; i <= top; i++) ok[a[stk[i]].id] = 1;
  for (int i = 1; i < top; i++) {
    if (a[stk[i]].b <= a[stk[i + 1]].b)
      ok[a[stk[i]].id] = 0;
    else
      break;
  }
  for (int i = 2; i <= n; i++) {
    if (a[i].k == a[i - 1].k && a[i].b == a[i - 1].b && ok[a[i - 1].id])
      ok[a[i].id] = 1;
  }
  for (int i = 1; i <= n; i++) {
    if (ok[i]) printf("%d ", i);
  }
  putchar('\n');
  return 0;
}
