Description

　　在xoy直角坐标平面上有n条直线L1,L2,...Ln,若在y值为正无穷大处往下看,能见到Li的某个子线段,则称Li为

可见的,否则Li为被覆盖的.

例如,对于直线:

L1:y=x; L2:y=-x; L3:y=

则L1和L2是可见的,L3是被覆盖的.

给出n条直线,表示成y=Ax+B的形式(|A|,|B|<=),且n条直线两两不重合.求出所有可见的直线.

Input

　　第一行为N( < N < ),接下来的N行输入Ai,Bi

Output

　　从小到大输出可见直线的编号，两两中间用空格隔开,最后一个数字后面也必须有个空格

Sample Input

- 

Sample Output
 

#include<iostream>

#include<cstdio>

#include<cmath>

#include<ctime>

#include<queue>

#include<algorithm>

#include<cstring>

using namespace std;

#define duke(i,a,n) for(int i = a;i <= n;i++)

#define lv(i,a,n) for(int i = a;i >= n;i--)

#define clean(a) memset(a,0,sizeof(a))

const int INF =  << ;

typedef long long ll;

typedef double db;

template <class T>

void read(T &x)

{

    char c;

    bool op = ;

    while(c = getchar(), c < '' || c > '')

        if(c == '-') op = ;

    x = c - '';

    while(c = getchar(), c >= '' && c <= '')

        x = x *  + c - '';

    if(op) x = -x;

}

template <class T>

void write(T x)

{

    if(x < ) putchar('-'), x = -x;

    if(x >= ) write(x / );

    putchar('' + x % );

}

const db eps = 1e-;

struct node

{

    db a,b;

    int n;

}l[];

int n,top;

bool ans[];

node st[];

bool cmp(node a,node b)

{

    if(fabs(a.a - b.a) < eps)

    return a.b < b.b;

    else

    return a.a < b.a;

}

db crossx(node x1,node x2)

{

    return (x2.b - x1.b) / (x1.a - x2.a);

}

void insert(node a)

{

    while(top)

    {

        if(fabs(st[top].a - a.a) < eps)

        top--;

        else if(top >  && crossx(a,st[top - ]) <= crossx(st[top],st[top - ]))

        top--;

        else

        break;

    }

    st[++top] = a;

}

void work()

{

    duke(i,,n)

    insert(l[i]);

    duke(i,,top)

    {

        ans[st[i].n] = ;

    }

    duke(i,,n)

    if(ans[i] == )

    printf("%d ",i);

}

int main()

{

    read(n);

    duke(i,,n)

    {

        scanf("%lf%lf",&l[i].a,&l[i].b);

        l[i].n = i;

    }

    sort(l + ,l + n + ,cmp);

    work();

    return ;

}