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;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= 1000000007;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long gcd(long long x, long long y) {
  while (y) {
    long long t = x % y;
    x = y;
    y = t;
  }
  return x;
}
long long inv(long long x) { return powmod(x, 1000000007 - 2); }
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, P;
struct point {
  int x, y, id;
  bool operator<(const point &p) const {
    return make_pair(x, y) > make_pair(p.x, p.y);
  }
} p[401000], s[401000];
int top;
set<pair<int, int> > res;
pair<int, int> b[401000];
map<int, int> mp;
bool cross(point a, point b, point c) {
  return 1LL * b.y * c.x * (a.x - b.x) * (a.y - c.y) <
         1LL * c.y * b.x * (a.y - b.y) * (a.x - c.x);
}
void solve() {
  scanf("%d", &n), m = n;
  int top = 0;
  for (int i = 0; i < n; i++) {
    p[i].x = read(), p[i].y = read();
    p[i].id = i + 1;
    b[i] = make_pair(p[i].x, p[i].y);
  }
  sort(p, p + n);
  for (int i = 0; i < n; i++) {
    if (!top || p[i].y > s[top - 1].y) s[top++] = p[i];
  }
  for (int i = 0; i < top; i++) p[i] = s[i];
  n = top, top = 0;
  for (int i = 0; i < n; i++) {
    while (top > 1 && cross(s[top - 2], s[top - 1], p[i])) top--;
    s[top++] = p[i];
  }
  for (int i = 0; i < top; i++) mp[s[i].x * 10000 + s[i].y] = 1;
  for (int i = 0; i < m; i++)
    if (mp[b[i].first * 10000 + b[i].second]) printf("%d ", i + 1);
}
int main() {
  solve();
  return 0;
}
