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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/priority_queue.hpp>
using namespace std;
using namespace __gnu_pbds;
#define REP(i,j)       for(int i = 0 ; i < j ; i++)
#define REPNM(i,j,k)   for(int i = j ; i < k ; i++)
#define RREP(i,j,k)    for(int i = j ; i >=k ; i--)
#define LREP(i,j)      for(LL  i = 0 ; i < j ; i++)
#define LREPNM(i,j,k)  for(LL  i = j ; i < k ; i++)
#define LRREP(i,j,k)   for(LL  i = j ; i >=k ; i--)
#define A    first
#define B    second
#define pb   push_back
#define mp   make_pair
#define LL  long long int
#define PII pair<int , int>
#define PLL pair< LL , LL >
#define VI  vector<int> 
#define VLL vector<LL> 
#define VPII vector<PII>
#define VPLL vector<PLL>
#define MEM(i,j)   memset(i , j , sizeof i)
#define ALL(i)     i.begin() , i.end()
#define DBGG(i,j)     cout << i << " " << j << endl
#define DB4(i,j,k,l)  cout << i << " " << j << " " << k << " " << l << endl
#define RI(i)       scanf("%d" , &i)
#define RII(i,j)    scanf("%d%d" , &i , &j)
#define RIII(i,j,k) scanf("%d%d%d" , &i , &j , &k)
#define RL(i)       scanf("%lld" , &i)
#define RLL(i,j)    scanf("%lld%lld" , &i , &j)
#define RLLL(i,j,k) scanf("%lld%lld%lld" , &i , &j , &k)
#define AS assert
#define IOS cin.tie() , cout.sync_with_stdio(0)
///------------------------------------------------------------
#define int long long
#define MAX 200090
#define INF 0x3f3f3f3f

typedef pair<PII , int> PIII;
PIII x[MAX] , p[MAX];
int n;
int cross(PII q , PII w , PII e){
    PII a = mp(w.A - q.A , w.B - q.B);
    PII b = mp(e.A - w.A , e.B - w.B);
    int tmp = a.A * b.B - a.B * b.A;
    if(tmp == 0) return 0;
    return (tmp > 0) ? 1 : -1;
}
vector<PIII> u , d;
int32_t main(){
    cin >> n;
    x[0].A = mp(0 , 0) , x[0].B = -1;
    REPNM(i , 1 , n + 1) cin >> x[i].A.A >> x[i].A.B , x[i].B = i;
    sort(x , x + n + 1);

    REP(i , n + 1){
        while(u.size() >= 2 && cross(u[u.size() - 2].A , u[u.size() - 1].A , x[i].A) == -1) u.pop_back();
        u.pb(x[i]);
    }
    RREP(i , n , 0){
        while(d.size() >= 2 && cross(d[d.size() - 2].A , d[d.size() - 1].A , x[i].A) == -1) d.pop_back();
        d.pb(x[i]);
    }

    set<PIII> ans;
    for(auto i : u) if(i.B > 0) ans.insert(i);
    for(auto i : d) if(i.B > 0) ans.insert(i);

    int tpo = 0;
    for(auto i : ans) p[tpo ++] = i;

   	n = ans.size();
   	// DBGG("n = " , n);
    sort(p , p + n);
    VI an;
    PII mm = p[n - 1].A;
    RREP(i , n - 1 , 0){
        PII now = p[i].A;
        int po = p[i].B;
        if(now.A == mm.A && now.B == mm.B) an.pb(po);
        if(now.A != mm.A && now.B > mm.B) mm = now , an.pb(po);
    }

    sort(ALL(an));
    for(auto i : an) cout << i << " " ; cout << endl;
    return 0;
}