(全国多校重现赛一) H Numbers

zk has n numbers a1,a2,...,ana1,a2,...,an. For each (i,j) satisfying 1≤i<j≤n, zk generates a new number (ai+aj)(ai+aj). These new numbers could make up a new sequence b1，b2,...,bn(n−1)/2b1，b2,...,bn(n−1)/2. 

LsF wants to make some trouble. While zk is sleeping, Lsf mixed up sequence a and b with random order so that zk can't figure out which numbers were in a or b. "I'm angry!", says zk. 

Can you help zk find out which n numbers were originally in a?

Input

Multiple test cases(not exceed 10). 

For each test case: 

∙∙The first line is an integer m(0≤m≤125250), indicating the total length of a and b. It's guaranteed m can be formed as n(n+1)/2. 

∙∙The second line contains m numbers, indicating the mixed sequence of a and b. 

Each aiai is in [1,10^9]

Output

For each test case, output two lines. 

The first line is an integer n, indicating the length of sequence a; 

The second line should contain n space-seprated integers a1,a2,...,an(a1≤a2≤...≤an)a1,a2,...,an(a1≤a2≤...≤an). These are numbers in sequence a. 

It's guaranteed that there is only one solution for each case.

Sample Input

6
2 2 2 4 4 4
21
1 2 3 3 4 4 5 5 5 6 6 6 7 7 7 8 8 9 9 10 11

Sample Output

3
2 2 2
6
1 2 3 4 5 6

这个看代码吧。QAQ

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int n,m;
vector <int> a,b,c,ans;
map <int,int> mmp;
int main()
{
	while(~scanf("%d",&n))
	{
		a.clear();
		b.clear();
		c.clear();
		ans.clear();
		mmp.clear();
		int temp;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&temp);
			a.push_back(temp);
			if(mmp[temp]==0) mmp[temp]=1;
			else mmp[temp]++; 

		}
		sort(a.begin(),a.end());
		ans.push_back(a[0]);
		for(int i=1;i<n;i++)
		{
			if(mmp[a[i]]==0) continue;
			for(int j=0;j<ans.size();j++)
			{
				mmp[ans[j]+a[i]]--;
			}
			ans.push_back(a[i]);
			mmp[a[i]]--;
		}

		printf("%d\n",ans.size());
		vector <int> ::iterator it;
		for(it=ans.begin();it!=ans.end();it++)
		{
			if(it==ans.begin()) cout<<*it;
			else cout<<" "<<*it;
		}
		cout<<endl;
	}
	return 0;
}