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 <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <iostream>
#include <iomanip>
#include <set>
#include <cmath>
#include <ctime>
#include <bitset>
using namespace std;

const int maxn = 300005;
const double eps = 1e-10;
int n;
struct poi  {
    double x, y;
    int num;
}p[maxn];

inline bool cmp(poi a, poi b)  {
    if (fabs(a.x-b.x)>eps) return a.x < b.x;
    if (fabs(a.y-b.y)>eps)return a.y<b.y;
    return a.num<b.num;
}

int ans[maxn];
vector<int> all[maxn];
void brute()  {
    int top=0;
    for (int i = 1; i <= n; i++)  {
        double minv = -1e9, maxv = 1e9;
        for (int j = i + 1; j <= n; j ++)  
            minv = max(minv, (p[i].y-p[j].y)/(p[i].x-p[j].x)*p[i].x*p[j].x/p[i].y/p[j].y);
        for (int j = i - 1; j > 0; j --)  {
            maxv = min(maxv, (p[i].y-p[j].y)/(p[i].x-p[j].x)*p[i].x*p[j].x/p[i].y/p[j].y);
            if (minv - eps > maxv) break;
        }
        if (minv - eps <= maxv)  {
            for (int j =0; j <(int)all[i].size();j++)
                ans[++top]=all[i][j];
        }
    }
    sort(ans+1,ans+1+top);
    for (int i =1;i <= top; i ++) printf("%d ", ans[i]);
}

vector<pair<int,int> >  yy[maxn];
int main()  {
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++) {
        int a,b;scanf("%d%d", &a,&b);
        yy[a].push_back(make_pair(b, i));
    }
    n=0;
    for (int i=1;i<=10000;i++)  {
        sort(yy[i].begin(),yy[i].end());
        if ((int)yy[i].size()==0)continue;
        int maxb=yy[i][(int)yy[i].size()-1].first;
        p[++ n] = (poi){i, maxb, i};
        all[n].clear();
        for (int j=0;j<(int)yy[i].size();j++)
            if (yy[i][j].first==maxb)  all[n].push_back(yy[i][j].second);
    }
    sort(p + 1, p + 1 + n, cmp);
    //printf("st\n");
    brute();
    return 0;
}