Codeforces 798D - Mike and distribution

D. Mike and distribution

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Mike has always been thinking about the harshness of social inequality. He's so obsessed with it that sometimes it even affects him while solving problems. At the moment, Mike has two sequences of positive integers A = [a1, a2, ..., an] and B = [b1, b2, ..., bn] of length neach which he uses to ask people some quite peculiar questions.

To test you on how good are you at spotting inequality in life, he wants you to find an "unfair" subset of the original sequence. To be more precise, he wants you to select k numbers P = [p1, p2, ..., pk] such that 1 ≤ pi ≤ n for 1 ≤ i ≤ k and elements in P are distinct. Sequence P will represent indices of elements that you'll select from both sequences. He calls such a subset P "unfair" if and only if the following conditions are satisfied: 2·(ap1 + ... + apk) is greater than the sum of all elements from sequence A, and 2·(bp1 + ... + bpk) is greater than the sum of all elements from the sequence B. Also, k should be smaller or equal to  because it will be to easy to find sequence P if he allowed you to select too many elements!

Mike guarantees you that a solution will always exist given the conditions described above, so please help him satisfy his curiosity!

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of elements in the sequences.

On the second line there are n space-separated integers a1, ..., an (1 ≤ ai ≤ 109) — elements of sequence A.

On the third line there are also n space-separated integers b1, ..., bn (1 ≤ bi ≤ 109) — elements of sequence B.

Output

On the first line output an integer k which represents the size of the found subset. k should be less or equal to .

On the next line print k integers p1, p2, ..., pk (1 ≤ pi ≤ n) — the elements of sequence P. You can print the numbers in any order you want. Elements in sequence P should be distinct.

Example

input

5
8 7 4 8 3
4 2 5 3 7

output

3
1 4 5
题目大意：略。
思路提示：2*(ap1 + ... + apk)>sum(ai)==>ap1+...+apk>ai中剩下的和。
方法：给a降序排序，b不排序。输出最大的a1的下标；然后从排好序的a2开始循环，增量为2。因为a1已经是最大的了，所以a2和a3无论选哪一个剩下的那一个都没a1大，所以选下标使b更大的那个，以此类推。。。。
所以，选出了a和b都更大的那一半。
代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=1e5+5;
struct node
{
    int n,id;
}a[N],b[N];
bool cmp(node a,node b)
{
    return a.n>b.n; 
} 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    cin>>a[i].n,a[i].id=i;
    for(int i=1;i<=n;i++)
    cin>>b[i].n,b[i].id=i;
    sort(a+1,a+n+1,cmp);
    cout<<n/2+1<<endl<<a[1].id;
    for(int i=2;i<=n;i+=2)
    {
        if(i<n)
        {
            if(b[a[i].id].n>=b[a[i+1].id].n)cout<<' '<<a[i].id;
            else cout<<' '<<a[i+1].id; 
        }
        else cout<<' '<<a[i].id;
    }
    cout<<endl;
    return 0;
}