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;

typedef long long ll;

struct frac{
    ll x, y;
    frac(ll a = 0, ll b = 0){
        //assert(b);
        if(b < 0) a = -a, b = -b;
        if(a){
            ll g = __gcd(abs(a), b);
            a /= g;
            b /= g;
        }
        else b = 1;
        x = a, y = b;
    }
    frac(ll a){
        x = a, y = 1;
    }

    void print(){
        //if(y == 1) printf("%lld\n", x);
        printf("%lld/%lld\n", x, y);
    }
    frac operator+(const frac &r) const{
        return frac(x * r.y + y * r.x, y * r.y);
    }
    frac operator-(const frac &r) const{
        return frac(x * r.y - y * r.x, y * r.y);
    }
    frac operator*(const frac &r) const{
        return frac(x * r.x, y * r.y);
    }
    frac operator/(const frac &r) const{
        return frac(x * r.y, y * r.x);
    }
    bool operator <(const frac &r) const{
        return x * r.y - y * r.x < 0;
    }

    bool operator >(const frac &r) const{
        return x * r.y - y * r.x > 0;
    }

    bool operator ==(const frac &r) const{
        return x * r.y - y * r.x == 0;
    }
};

const int maxn = 4e5 + 233;
frac x[maxn], y[maxn];
int ord[maxn];
int n;

int s[maxn];
int top;

bool Cross(frac x1, frac y1, frac x2, frac y2){
    return x1 * y2 - x2 * y1 > frac(0, 1);
}

int main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i += 1){
        int a, b;
        cin >> a >> b;
        x[i] = frac(1, a);
        y[i] = frac(1, b);
        ord[i] = i;
    }

    sort(ord + 1, ord + n + 1, [&](int a, int b){
        if(x[a] == x[b]) return y[a] < y[b];
        return x[a] < x[b];
    });

    int cnt = 0;
    for (int i = 1; i <= n; i++){
        int j = i;
        while(j <= n && x[ord[j]] == x[ord[i]]) j++;
        for(int k = i; k < j; k += 1){
            if(y[ord[k]] == y[ord[i]]) {
                ord[++cnt] = ord[k];
            }
            else break;
        }
        i = j - 1;
    }

    for(int o = 1; o <= cnt; o += 1){
        int i = ord[o];
        while(top > 1 && Cross(x[i] - x[s[top]], y[i] - y[s[top]], 
            x[s[top]] - x[s[top - 1]], y[s[top]] - y[s[top - 1]])){
            top--;
        }
        s[++top] = i;
    }

    vector<int> res;
    

    res.emplace_back(s[1]);
    for (int i = 2; i <= top; i += 1){
        if(y[s[i]] > y[s[i - 1]] || 
            (x[s[i]] > x[s[i - 1]] && 
                y[s[i]] == y[s[i - 1]])) break;
        res.emplace_back(s[i]);
    }
    sort(res.begin(), res.end());
    for(int i = 0; i < res.size(); i++){
        cout << res[i] << " \n"[i == res.size() - 1];
    }

    if(n == 200000){
        x[14034].print();
        y[14034].print();
        x[28469].print();
        y[28469].print();
        x[51162].print();
        y[51162].print();
        x[105995].print();
        y[105995].print();
        x[106984].print();
        y[106984].print();
        x[136414].print();
        y[136414].print();
        x[175766].print();
        y[175766].print();
    }
    return 0;
}
