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 node {
  int num;
  int x, y;
} a[200005], c[200005], b[200005];
vector<int> ret;
int n;
int m;
bool vis[200005];
bool cmp(node a, node b) {
  if (a.x != b.x) return a.x < b.x;
  return a.y > b.y;
}
bool judge(node p0, node p1, node p2) {
  int k1 = p2.y, k2 = p1.y, k3 = p0.y, k4 = p2.x, k5 = p1.x, k6 = p0.x;
  return 1ll * (k1 * k3 - k2 * k3) * (k4 * k6 - k4 * k5) >
         1ll * (k1 * k2 - k1 * k3) * (k5 * k6 - k4 * k6);
}
void tubao() {
  m = 1;
  c[1] = a[1];
  for (int i = 2; i <= n; i++) {
    if (a[i].x == a[i - 1].x) continue;
    while (m >= 2 && judge(a[i], c[m], c[m - 1])) m--;
    c[++m] = a[i];
  }
  for (int i = 1; i <= m; i++) vis[c[i].num] = 1;
  for (int i = 2; i <= m; i++) {
    if (c[i - 1].y <= c[i].y) {
      vis[c[i - 1].num] = 0;
    } else
      break;
  }
  for (int i = 2; i <= n; i++) {
    if (a[i].y == a[i - 1].y && a[i].x == a[i - 1].x && vis[a[i - 1].num])
      vis[a[i].num] = true;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
    a[i].num = i;
  }
  sort(a + 1, a + n + 1, cmp);
  tubao();
  for (int i = 1; i <= n; i++) {
    if (vis[i]) printf("%d ", i);
  }
  return 0;
}
