[数论] Codeforces Round #324 (Div. 2)D. Dima and Lisa

Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at most than three primes.

More formally, you are given an odd numer n. Find a set of numbers pi (1 ≤ i ≤ k), such that

1. 1 ≤ k ≤ 3
2. pi is a prime
3. 

The numbers pi do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.

Input

The single line contains an odd number n (3 ≤ n < 109).

Output

In the first line print k (1 ≤ k ≤ 3), showing how many numbers are in the representation you found.

In the second line print numbers pi in any order. If there are multiple possible solutions, you can print any of them.

Sample test(s)

input

27

output

3
5 11 11

Note

A prime is an integer strictly larger than one that is divisible only by one and by itself.

数论：哥德巴赫猜想

任一大于2的偶数都可写成两个质数之和。

任一大于7的奇数都可写成三个素数之和。

如果n-2是素数，那么可以拆成2，n-2。否则拆成3个数 

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <fstream>
#include <math.h>
#include <iomanip>
#define read freopen("q.in","r",stdin)
using namespace std;
typedef long long LL;
#define INF 0x3fffffff
typedef pair<int, int> pii;
const int maxn = 300001;
const LL mod = 1000000007;

bool isp(int x)
{
    if(x==1)return false;
    for(int i=2;i*i<=x;i++)
        if(x%i==0)return false;
    return true;
}

int main()
{
    int n,i,j;
    while(~scanf("%d",&n))
    {
        int k=n-2;
        if(isp(n))printf("1\n%d\n",n);
        else if(isp(k))printf("2\n2 %d\n",k);
        else
        {
            int flag=0;
            for(i=3;i<=n && !flag;i+=2)
            {
                for(j=3;j<=n;j+=2)
                {
                    if(isp(i) && isp(j) && isp(n-i-j))
                    {
                        printf("3\n%d %d %d\n",i,j,n-i-j);
                        flag=1;
                        break;
                    }
                }
            }
        }

    }
}