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;
void ga(int N, int* A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
struct pt {
  long long x, y, I;
  pt operator-(pt r) { return {x - r.x, y - r.y}; }
  bool operator<(const pt& r) const { return (x == r.x) ? y > r.y : x < r.x; }
  long long XP(pt b) { return x * b.y - y * b.x; }
  bool gt() { return ~scanf("%lld%lld", &x, &y); }
  void prt() { printf("%lld %lld\n", x, y); }
  bool qq(pt a) { return x ^ a.x && y ^ a.y; }
} p[(1 << 21)], r[(1 << 21)];
bool test(pt& a, pt& b, pt& c) {
  if (a.x == b.x && a.x == c.x) return 1;
  if (a.y == b.y && a.y == c.y) return 1;
  return 0;
}
set<pt> T;
long long XX(pt& a, pt& b, pt& c) {
  return (a.x - c.x) * (b.y - c.y) - (b.x - c.x) * (a.y - c.y);
}
bool cp(pt a, pt b) { return (a.y == b.y) ? a.x < b.x : a.y < b.y; }
int hAW(pt* p, int n, pt* r) {
  sort(p, p + n);
  int m = 0;
  for (int i(0); i < n; i++) {
    while (m >= 2 && (XX(r[m - 2], r[m - 1], p[i]) < 0 or
                      (XX(r[m - 2], r[m - 1], p[i]) == 0 &&
                       test(r[m - 2], r[m - 1], p[i]))))
      --m;
    r[m++] = p[i];
  }
  for (int i = n - 1, t = m + 1; ~i; --i) {
    while (m >= t && (XX(r[m - 2], r[m - 1], p[i]) < 0 or
                      (XX(r[m - 2], r[m - 1], p[i]) == 0 &&
                       test(r[m - 2], r[m - 1], p[i]))))
      --m;
    r[m++] = p[i];
  }
  return m - 1;
}
long long N, X, Y, C[(1 << 21)], M, S, x, y;
int main(void) {
  scanf("%lld", &N);
  for (int i(0); i < N; i++)
    p[i].gt(), p[i].I = i + 1, X = max(X, p[i].x), Y = max(Y, p[i].y);
  for (int i(0); i < N; i++)
    if (p[i].x == X) y = max(y, p[i].y);
  for (int i(0); i < N; i++)
    if (p[i].x == X && p[i].y ^ y) p[i].I = -1;
  for (int i(0); i < N; i++)
    if (p[i].y == Y) x = max(x, p[i].x);
  for (int i(0); i < N; i++)
    if (p[i].y == Y && p[i].x ^ x) p[i].I = -1;
  p[N++] = {-1, -1, -1}, p[N++] = {-1, Y - 1, -1}, p[N++] = {X - 1, -1, -1};
  M = hAW(p, N, r);
  for (int i(0); i < M; i++) T.insert(r[i]);
  for (int i(0); i < N; i++)
    if (~p[i].I && T.count(p[i])) ++S, ++C[p[i].I];
  for (int i(0); i < N + 2; i++)
    if (C[i]) printf("%d%c", i, --S ? 32 : 10);
  return 0;
}
