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;
struct xd {
  double x, y;
  int idx;
  inline bool operator<(const xd& b) const {
    return x != b.x ? x < b.x : y < b.y;
  }
  inline bool operator==(const xd& b) const { return x == b.x && y == b.y; }
} all[200005];
int stk[200005], top;
bool ans[200005];
double cross_z(xd o, xd a, xd b) {
  return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x);
}
int main() {
  int N;
  while (~scanf("%d", &(N))) {
    int x, y;
    for (int i = 0; i < (N); ++i) {
      scanf("%d%d", &(x), &(y));
      all[i].x = 1. / x, all[i].y = 1. / y;
      all[i].idx = i;
    }
    sort(all, all + N);
    top = 0;
    for (int i = 0; i < (N); ++i) {
      if (i == 0 || all[i].x != all[i - 1].x) {
        while (top >= 2 &&
               cross_z(all[stk[top - 2]], all[stk[top - 1]], all[i]) < -1e-20)
          --top;
        stk[top++] = i;
      }
    }
    while (top >= 2 && all[stk[top - 2]].y <= all[stk[top - 1]].y) --top;
    memset(ans, false, sizeof(ans));
    for (int i = 0; i < (top); ++i) ans[all[stk[i]].idx] = true;
    for (int i = 1; i < N; ++i)
      if (ans[all[i - 1].idx] && all[i] == all[i - 1]) ans[all[i].idx] = true;
    for (int i = 0; i < (N); ++i)
      if (ans[i]) printf("%d ", i + 1);
    putchar('\n');
  }
  return 0;
}
