Tavas is a cheerleader in the new sports competition named "Pashmaks".

<image>

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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int maxN = 210000;
int s[maxN], r[maxN];
int order[maxN];
int n;
int mx[maxN];
bool comp(int x, int y) {
  return make_pair(s[x], r[x]) < make_pair(s[y], r[y]);
}
double getValue(int i, int j) {
  double A = (double)(r[i] - r[j]) / (double)(r[i] * r[j]);
  double B = (double)(s[i] - s[j]) / (double)(s[i] * s[j]);
  if (s[i] == s[j]) {
    if (r[i] > r[j]) {
      return 1e50;
    } else if (r[i] < r[j]) {
      return -1e50;
    } else {
      return 0;
    }
  }
  return A / B;
}
double getMax(vector<int> &candidates, int pos) {
  if (pos + 1 == candidates.size()) {
    return -1e50;
  }
  int left_bound = pos + 1, right_bound = candidates.size() - 1;
  while (right_bound - left_bound > 10) {
    int m1 = (2 * left_bound + right_bound) / 3;
    int m2 = (left_bound + 2 * right_bound) / 3;
    double v1 = getValue(candidates[pos], candidates[m1]);
    double v2 = getValue(candidates[pos], candidates[m2]);
    if (v1 > v2) {
      right_bound = m2;
    } else {
      left_bound = m1;
    }
  }
  double res = -1e50;
  for (int i = left_bound; i <= right_bound; ++i) {
    double v = getValue(candidates[pos], candidates[i]);
    if (v > 1e40) {
      continue;
    }
    res = max(res, v);
  }
  return res;
}
double getMin(vector<int> &candidates, int pos) {
  if (pos == 0) {
    return 1e50;
  }
  int left_bound = 0, right_bound = pos - 1;
  while (right_bound - left_bound > 10) {
    int m1 = (2 * left_bound + right_bound) / 3;
    int m2 = (left_bound + 2 * right_bound) / 3;
    double v1 = getValue(candidates[pos], candidates[m1]);
    double v2 = getValue(candidates[pos], candidates[m2]);
    if (v1 < v2) {
      right_bound = m2;
    } else {
      left_bound = m1;
    }
  }
  double res = 1e50;
  for (int i = left_bound; i <= right_bound; ++i) {
    double v = getValue(candidates[pos], candidates[i]);
    res = min(res, v);
  }
  return res;
}
void gen() {
  int n = 200000;
  printf("%d\n", n);
  for (int i = 0; i < n; ++i) {
    printf("%d %d\n", rand() % 10000 + 1, rand() % 10000 + 1);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &s[i], &r[i]);
    order[i] = i;
  }
  sort(order, order + n, comp);
  for (int i = n - 1; i >= 0; --i) {
    mx[i] = r[order[i]];
    if (i + 1 < n) {
      mx[i] = max(mx[i], mx[i + 1]);
    }
  }
  vector<int> candidates;
  for (int i = 0; i < n; ++i) {
    if (i + 1 == n || r[order[i]] > mx[i + 1]) {
      candidates.push_back(order[i]);
    }
  }
  vector<int> res;
  set<pair<int, int> > S;
  for (int i = 0; i < candidates.size(); ++i) {
    double A = getMax(candidates, i);
    double B = getMin(candidates, i);
    if (A <= B + 1e-9 && A < 1e-9) {
      S.insert(make_pair(s[candidates[i]], r[candidates[i]]));
    }
  }
  for (int i = 0; i < n; ++i) {
    if (S.count(make_pair(s[i], r[i])) > 0) {
      printf("%d ", i + 1);
    }
  }
  printf("\n");
  return 0;
}
