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;
template <class Int>
inline void read(Int &x) {
  x = 0;
  char ch = getchar();
  bool fl = 0;
  for (; ch > '9' || ch < '0'; ch = getchar())
    if (ch == '-') fl = 1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + (ch - 48);
  if (fl) x = -x;
}
const int N = 3e5 + 10;
const double INF = 1e18;
const double eps = 1e-9;
int n;
struct neos {
  int a, b;
  int id;
  neos() {}
  neos(int _a, int _b, int i) { a = _a, b = _b, id = i; }
  inline bool operator<(neos B) const {
    if (a != B.a) return a < B.a;
    return b < B.b;
  }
} G[N], D[N];
inline double calc(double ai, double bi, double aj, double bj) {
  return (ai * aj / (bi * bj)) * (bi - bj) / (aj - ai);
}
int st[N], top = 0, cnt = 0;
int vis[N];
bool ok[N];
int main() {
  read(n);
  int a, b;
  for (int i = 1; i <= n; ++i) {
    read(a);
    read(b);
    G[i] = neos(a, b, i);
  }
  sort(G + 1, G + 1 + n);
  int maxb = -1, lst = n;
  for (int i = n; i; --i) {
    if (G[i].b > maxb) {
      D[++cnt] = G[i];
      lst = i;
      maxb = G[i].b;
    } else if (G[i].b == maxb) {
      if (G[i].a == G[lst].a) vis[G[i].id] = G[lst].id;
    }
  }
  if (cnt <= 5000) {
    for (int i = 1; i <= cnt; ++i) {
      double dlim = eps, ulim = INF;
      bool fl = 1;
      for (int j = 1; j <= cnt; ++j) {
        if (j == i) continue;
        if (D[i].a == D[j].a) {
          if (D[j].b > D[i].b) {
            fl = 0;
            break;
          }
          continue;
        }
        double X = calc(D[i].a, D[i].b, D[j].a, D[j].b);
        if (D[i].a < D[j].a) {
          if (X < dlim) {
            fl = 0;
            break;
          }
          ulim = min(ulim, X);
        } else {
          if (X > ulim) {
            fl = 0;
            break;
          }
          dlim = max(dlim, X);
        }
        if (ulim < dlim) {
          fl = 0;
          break;
        }
      }
      if (fl) st[++top] = D[i].id, ok[D[i].id] = 1;
    }
    for (int i = 1; i <= n; ++i)
      if (vis[i] && ok[vis[i]]) st[++top] = i;
    sort(st + 1, st + 1 + top);
    for (int i = 1; i <= top; ++i) printf("%d ", st[i]);
    puts("");
  } else {
    top = 0;
    for (int i = 1; i <= cnt; ++i) {
      while (top > 1 && calc(D[st[top]].a, D[st[top]].b, D[i].a, D[i].b) >
                            calc(D[st[top]].a, D[st[top]].b, D[st[top - 1]].a,
                                 D[st[top - 1]].b))
        --top;
      st[++top] = i;
    }
    for (int i = 1; i <= top; ++i) st[i] = D[st[i]].id, ok[st[i]] = 1;
    for (int i = 1; i <= n; ++i)
      if (vis[i] && ok[vis[i]]) st[++top] = i;
    sort(st + 1, st + 1 + top);
    for (int i = 1; i <= top; ++i) printf("%d ", st[i]);
    puts("");
  }
  return 0;
}
