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 A, class B>
inline bool mina(A &x, B y) {
  return (x > y) ? (x = y, 1) : 0;
}
template <class A, class B>
inline bool maxa(A &x, B y) {
  return (x < y) ? (x = y, 1) : 0;
}
int X[(200005)], Y[(200005)];
int N;
set<tuple<int, int> > good;
int maxY[(10005)];
int sz_S;
tuple<int, int> S[(200005)];
void add_point(int x3, int y3) {
  while (sz_S >= 2) {
    int x1 = get<0>(S[sz_S - 2]), y1 = get<1>(S[sz_S - 2]);
    int x2 = get<0>(S[sz_S - 1]), y2 = get<1>(S[sz_S - 1]);
    if ((long long)y2 * x3 * (x1 - x2) * (y1 - y3) <
        (long long)x2 * y3 * (y1 - y2) * (x1 - x3)) {
      sz_S--;
    } else {
      break;
    }
  }
  S[sz_S++] = tuple<int, int>(x3, y3);
}
int main() {
  scanf("%d", &N);
  for (int(i) = 0; (i) < (N); ++(i)) {
    scanf("%d %d", &X[i], &Y[i]);
    maxa(maxY[X[i]], Y[i]);
  }
  int pre_y = 0;
  for (int i = (10005) - 1; i >= 1; i--) {
    if (maxY[i] > pre_y) {
      add_point(i, maxY[i]);
      pre_y = maxY[i];
    }
  }
  for (int(i) = 0; (i) < (sz_S); ++(i)) {
    good.insert(S[i]);
  }
  for (int(i) = 0; (i) < (N); ++(i)) {
    if (good.find(tuple<int, int>(X[i], Y[i])) != good.end()) {
      printf("%d ", i + 1);
    }
  }
  return 0;
}
