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 <stdio.h>
# include <bits/stdc++.h>
using namespace std;
# define fi cin
# define fo cout
# define x first
# define y second
# define ll long long
# define IOS ios_base :: sync_with_stdio(0);cin.tie(0)
# define p(v) cerr << #v << " = " << v << '\n'
# define p2(v) cerr << #v << " = " << (complex < db > (v.x,v.y)) << '\n'
# define vi vector < int >
# define vll vector < ll >
# define pii pair < int , int >
# define mp make_pair
# define db long double
# define pdd pair < db , db >
bool operator == (pdd a,pdd b)
{
    return abs(a.x - b.x) < 1e-18 && abs(a.y - b.y) < 1e-18;
}
int main(void)
{
    #ifdef CF
    freopen("input","r",stdin);
    #endif // CF
    srand(time(0));
    fo << fixed << setprecision(7);
    cerr << fixed << setprecision(7);
    int n;
    IOS;
    int a,b;
    fi>>n;
    static pdd s[1 << 20];
    static int p[1 << 20];
    for (int i = 1;i <= n;++i)
        fi>>a>>b,p[i] = i,s[i].x = 1.0 / a,s[i].y = 1.0 / b;
    sort(p + 1,p + 1 + n,[&](int a,int b)
         {
             return s[a] < s[b];
         });
    static vi h;
    auto inter = [&](pdd a,pdd b,pdd c)
    {
        return (c.x - b.x) * (a.y - c.y) - 1e-18 < (c.x - a.x) * (b.y - c.y);
    };
    for (int i = 1;i <= n;++i)
    {
        while (h.size() >= 2 && inter(s[h[h.size() - 2]],s[h[h.size() - 1]],s[p[i]]))
            h.pop_back();
        if (h.empty() || (!h.empty() && s[h.back()] != s[p[i]]))
            h.push_back(p[i]);
    }
    auto sign = [&](auto x)
    {
        return x < 0 ? -1 : 1;
    };
    static int ok[1 << 20];
    ok[h[0]] = 1;
    for (int i = 1;i < h.size();++i)
        if (sign(s[h[i - 1]].x - s[h[i]].x) == sign(s[h[i]].y - s[h[i - 1]].y) && s[h[i]].x != s[h[i - 1]].x)
            ok[h[i]] = 1;
    for (int i = n;i;--i)
    {
        while (!h.empty() && h.back() > i) h.pop_back();
        if (!h.empty() && s[h.back()] == s[i] && ok[h.back()])
            ok[i] = 1;
    }
    for (int i = 1;i <= n;++i)
        if (ok[i])
            fo << i << ' ';
    fo << '\n';
    cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms" << '\n';
    return 0;
}


