Codeforces Round #410 (Div. 2) D. Mike and distribution 思维

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. SequenceP 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

题意是给两个序列，让你选几个下标，使得对于这两个序列都满足所选的数和大于未选的数和

思维水题，先按a值从大到小排序，扫一遍，先取最大的那个值，然后两个两个的选，优先选b值较大的，如果总数为偶数个，则最后还会剩一个，选上就好了，这样不难想ab都会满足那个条件。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define eps 1e-9
#define PI 3.141592653589793
#define bs 1000000007
#define bsize 256
#define MEM(a) memset(a,0,sizeof(a))
typedef long long ll;
using namespace std;
struct node
{
	int a,b,index;
}num[100005];
int cmp(node a,node b)
{
	return a.a>b.a;
}
int main()
{
	int n,i,j;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	scanf("%d",&num[i].a),num[i].index=i;
	for(i=1;i<=n;i++)
	scanf("%d",&num[i].b);
	sort(num+1,num+n+1,cmp);
	vector<int>ans;
	ans.push_back(num[1].index);
	for(i=2;i<=n;i+=2)
	{
		if(i+1<=n&&num[i+1].b>num[i].b)
		ans.push_back(num[i+1].index);
		else
		ans.push_back(num[i].index);
	}
	printf("%d\n",ans.size());
	for(i=0;i<ans.size();i++)
	printf("%d%c",ans[i]," \n"[i==n]);
	return 0;
 }