题目连接：

http://www.codeforces.com/contest/665/problem/D

Description

A tuple of positive integers {x1, x2, ..., xk} is called simple if for all pairs of positive integers (i,  j) (1  ≤ i  <  j ≤ k), xi  +  xj is a prime.

You are given an array a with n positive integers a1,  a2,  ...,  an (not necessary distinct). You want to find a simple subset of the array a with the maximum size.

A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's define a subset of the array a as a tuple that can be obtained from a by removing some (possibly all) elements of it.

Input

The first line contains integer n (1 ≤ n ≤ 1000) — the number of integers in the array a.

The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.

Output

On the first line print integer m — the maximum possible size of simple subset of a.

On the second line print m integers bl — the elements of the simple subset of the array a with the maximum size.

If there is more than one solution you can print any of them. You can print the elements of the subset in any order.

Sample Input

2
2 3

Sample Output

2
3 2

Hint

题意

给你n个数，你需要找到一个最大的子集，使得这个子集中的任何两个数加起来都是质数。

题解：

无视掉1的话，我们最多选择一个奇数和一个偶数，因为奇数+奇数=偶数，偶数加偶数=偶数
所以直接暴力枚举就好了。
另外这道题如果建边的话，跑dfs直接莽一波最大团也可以……

我是一个智障，我就跑了最大团 QAQ

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int pri[2000005];
bool flag[maxn], a[maxn][maxn];
int ans, cnt[maxn], group[maxn], n, vis[maxn];
// 最大团： V中取K个顶点，两点间相互连接
// 最大独立集： V中取K个顶点，两点间不连接
// 最大团数量 = 补图中最大独立集数bool dfs( int u, int pos ){int i, j;for( i = u+1; i <= n; i++){if( cnt[i]+pos <= ans ) return 0;if( a[u][i] ){// 与目前团中元素比较，取 Non-N(i)for( j = 0; j < pos; j++ ) if( !a[i][ vis[j] ] ) break;if( j == pos ){     // 若为空，则皆与 i 相邻，则此时将i加入到 最大团中vis[pos] = i;if( dfs( i, pos+1 ) ) return 1;}}}if( pos > ans ){for( i = 0; i < pos; i++ )group[i] = vis[i]; // 最大团 元素ans = pos;return 1;}return 0;
}
void maxclique()
{ans=-1;for(int i=n;i>0;i--){vis[0]=i;dfs(i,1);cnt[i]=ans;}
}
void pre()
{pri[1]=1;pri[0]=1;for(int i=2;i<2000005;i++){if(pri[i])continue;for(int j=i+i;j<2000005;j+=i)pri[j]=1;}
}
int aa[maxn];
int main()
{pre();scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&aa[i]);for(int i=1;i<=n;i++)for(int j=1;j<i;j++)if(!pri[aa[i]+aa[j]])a[i][j]=1,a[j][i]=1;maxclique();cout<<ans<<endl;for(int i=0;i<ans;i++)cout<<aa[group[i]]<<" ";cout<<endl;
}