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;

#ifdef Adrian
#include "debug.h"
#else
#define debug(...) 9999
#endif

typedef long long ll;
typedef long double ld;
//typedef complex<ld> point;
#define F first
#define S second
#define ii pair<ll,ll>
template<typename G1, typename G2 = G1, typename G3 = G1>
struct triple{ G1 F; G2 S; G3 T; };
typedef triple<ll> iii;

struct fraction
{
    ll x, y;
    fraction(ll _x, ll _y) : x(_x), y(_y)
    {
        ll g = __gcd(abs(x), abs(y));
        if(y < 0) x = -x, y = -y;
        x /= g;
        y /= g;
    }
    fraction(ll _x) : x(_x), y(1){}
    fraction operator +(const fraction o)const
    {
        return {x * o.y + y * o.x, y * o.y};
    }
    fraction operator -(const fraction o)const
    {
        return {x * o.y - y * o.x, y * o.y};
    }
    fraction operator *(const fraction o)const
    {
        return {x * o.x, y * o.y};
    }
    fraction operator *(const ll o)const
    {
        return {x * o, y};
    }
    bool operator <(const fraction o)const
    {
        return x * o.y < y * o.x;
    }
    bool operator >(const fraction o)const
    {
        return x * o.y > y * o.x;
    }
    bool operator ==(const fraction o)const
    {
        return x * o.y == y * o.x;
    }
};

struct point
{
    fraction F, S;
    int id;
    fraction operator *(const point o)const
    {
        return F * o.F + S * o.S;
    }
    fraction operator /(const point o)const
    {
        return F * o.S - S * o.F;
    }
    point operator +(const point o)const
    {
        return {F + o.F, S + o.S};
    }
    point operator -(const point o)const
    {
        return {F - o.F, S - o.S};
    }
    bool operator <(const point o)const
    {
        if(F == o.F && S == o.S) return id < o.id;
        if(F == o.F) return S > o.S;
        return F < o.F;
    }
};

int main()
{
#ifdef Adrian
    //freopen("test.txt", "r", stdin);
    //freopen("test.out", "w", stdout);
#else
    ios_base::sync_with_stdio(0); cin.tie(0);
#endif


    int n;
    cin>>n;

    vector<vector<int>> mp1(n);
    map<ii, int> mp2;
    vector<point> v;
    int c = 0;
    for(int i=0; i<n; i++)
    {
        int x, y;
        cin>>x>>y;
        if(mp2.count({x, y}) == 0)
        {
            v.push_back({{1, x}, {1, y}, c});
            mp2[{x, y}] = c++;
        }
        mp1[mp2[{x, y}]].push_back(i + 1);
    }

    sort(v.begin(), v.end());

    vector<point> hull;
    for(int i=0; i<c; i++)
    {
        while(hull.size() > 1 && (hull.end()[-1] - hull.end()[-2]) / (v[i] - hull.end()[-1]) < 0) hull.pop_back();
        if(hull.empty() || hull.back() < v[i])
            hull.push_back(v[i]);
    }

    vector<int> ans;
    for(auto &i:hull)
        for(auto j:mp1[i.id])
            ans.push_back(j);

    sort(ans.begin(), ans.end());
    for(int i=0; i<ans.size(); i++)
        cout<<ans[i]<<" \n"[i == ans.size() - 1];

    return 0;
}
