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 
5
10 50
10 50
10 50
10 50
10 50
 2
10000 10000
10000 10000
 3
10000 10000
10000 10000
10000 10000
 3
1000 3000
1500 1500
3000 1000
 43
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 1
1 1
 2
1 1
1 1
 44
78 52
75 16
70 7
75 16
54 2
70 7
54 2
70 7
54 2
70 7
54 2
54 2
70 7
70 7
54 2
70 7
75 16
70 7
54 2
54 2
54 2
54 2
54 2
54 2
75 16
70 7
70 7
54 2
70 7
75 16
54 2
75 16
75 16
54 2
54 2
70 7
70 7
75 16
70 7
70 7
70 7
75 16
75 16
54 2
 3
1000 3000
1499 1499
3000 1000
 5
50 50
49 50
50 49
49 49
50 1
 18
82 38
33 69
33 69
33 69
33 69
33 69
82 38
33 69
33 69
33 69
82 38
33 69
33 69
82 38
82 38
33 69
33 69
33 69
 1
10000 9999
 3
1000 3000
1501 1501
3000 1000
 2
1 2
2 1

1 2 3 4 5
 1 2
 1 2 3
 1 2 3
 1 3 16 24 29 31 32 37 41 42
 1
 1 2
 1
 1 3
 1
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
 1
 1 2 3
 1 2 

#include <bits/stdc++.h>
using namespace std;
int n, m, ans, tot;
int p = 31;
long long gcd(long long x, long long y) {
  while (y) {
    long long r = x % y;
    x = y;
    y = r;
  }
  return x;
}
long long qpow(long long a, long long b, long long modp) {
  long long ans;
  for (ans = 1; b; b >>= 1, a = a * a % modp)
    if (b & 1) ans = ans * a % modp;
  return ans;
}
long long getc(long long n, long long m) {
  if (n < m) return 0;
  if (m > n - m) m = n - m;
  long long s1 = 1, s2 = 1;
  for (long long i = 0; i < m; i++) {
    s1 = s1 * (n - i) % p;
    s2 = s2 * (i + 1) % p;
  }
  return s1 * qpow(s2, p - 2, p) % p;
}
long long fact[20];
void extgcd(long long a, long long b, long long &x, long long &y) {
  if (b != 0) {
    extgcd(b, a % b, y, x);
    y -= (a / b) * x;
  } else {
    x = 1;
    y = 0;
  }
}
long long mod_inv(long long a, long long modp) {
  long long x, y;
  extgcd(a, modp, x, y);
  return (modp + x % modp) % modp;
}
long long mod_fact(long long n, long long p, long long &e) {
  e = 0;
  if (n == 0) return 1;
  long long res = mod_fact(n / p, p, e);
  e += n / p;
  if (n / p % 2 != 0) return res * (p - fact[n % p]) % p;
  return res * fact[n % p] % p;
}
long long lucas(long long n, long long k) {
  if (n < 0 || k < 0 || n < k) return 0;
  long long e1, e2, e3;
  long long a1 = mod_fact(n, p, e1), a2 = mod_fact(k, p, e2),
            a3 = mod_fact(n - k, p, e3);
  if (e1 > e2 + e3) return 0;
  return a1 * mod_inv(a2 * a3 % p, p) % p;
}
const long double eps = 1e-14;
const long double pi = std::acos(-1.0);
pair<pair<int, int>, int> q[200007];
int v[200007], top;
pair<pair<int, int>, int> st[200007], st2[200007];
long double L[200007], R[200007];
long double calc(pair<pair<int, int>, int> &x, pair<pair<int, int>, int> &y) {
  long double s1 = x.first.first;
  long double r1 = x.first.second;
  long double s2 = y.first.first;
  long double r2 = y.first.second;
  long double g = s1 * s2 * r1 * r2;
  long double k = (g / s1 - g / s2) / (g / r1 - g / r2);
  return k;
}
int chk(pair<pair<int, int>, int> &x, pair<pair<int, int>, int> &y,
        pair<pair<int, int>, int> &z) {
  long double xx = calc(x, y);
  long double yy = calc(z, y);
  if (fabs(xx - yy) < eps) return 0;
  if (xx < yy)
    return -1;
  else
    return 1;
}
int main() {
  int T = 1;
  while (scanf("%d", &n) != EOF) {
    tot = 0;
    if (!n) break;
    for (int i = 0; i < n; ++i) {
      int x, y;
      scanf("%d%d", &x, &y);
      q[i].first = pair<int, int>(x, y);
      q[i].second = i + 1;
    }
    sort(q, q + n);
    m = n;
    for (int i = 0; i < n; ++i) {
      while (tot > 0 && (q[i].first.second > st[tot - 1].first.second ||
                         q[i].first.first > st[tot - 1].first.first &&
                             q[i].first.second == st[tot - 1].first.second)) {
        --tot;
        v[st[tot].second] = 1;
        --m;
      }
      st[tot++] = q[i];
    }
    top = 0;
    for (int i = 0; i < tot; ++i) {
      while ((top > 0 && st[i].first == st2[top - 1].first ||
              top > 1 && chk(st[i], st2[top - 1], st2[top - 2]) < 0))
        --top;
      st2[top++] = st[i];
      if (top > 1)
        L[i] = calc(st[i], st2[top - 2]);
      else
        L[i] = -((1 << 30) - 1);
    }
    top = 0;
    for (int i = tot - 1; i >= 0; --i) {
      while ((top > 0 && st[i].first == st2[top - 1].first ||
              top > 1 && chk(st[i], st2[top - 1], st2[top - 2]) > 0))
        --top;
      st2[top++] = st[i];
      if (top > 1)
        R[i] = calc(st[i], st2[top - 2]);
      else
        R[i] = 0;
    }
    for (int i = 0; i < tot; ++i) {
      if (!(L[i] < R[i] || fabs(L[i] - R[i]) < eps)) v[st[i].second] = 1, --m;
    }
    for (int i = 1, cnt = 0; i <= n; ++i) {
      if (!v[i]) {
        ++cnt;
        printf("%d%c", i, " \n"[cnt == m]);
      }
    }
  }
  return 0;
}
