Dima and Lisa（哥德巴赫猜想）

Problem description

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 p_i (1 ≤ i ≤ k), such that

1. 1 ≤ k ≤ 3
2. p_i is a prime
3. $sum_{i=1}^k p_i = n $

The numbers p_i 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 p_i in any order. If there are multiple possible solutions, you can print any of them.

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.

解题思路：将一个奇数拆分成1~3个素数，暴力即过！

哥德巴赫猜想：随便取某一个奇数，比如77，可以把它写成三个素数之和，即77=53+17+7；再任取一个奇数，比如461，可以表示成461=449+7+5，也是三个素数之和，461还可以写成257+199+5，仍然是三个素数之和。例子多了，即发现“任何大于5的奇数都是三个素数之和。”当然此题只是要求可以拆分成1~3个素数，并不要求一定是3个素数，譬如461本身就是素数，则此时直接输出1 461即可。

AC代码（31ms）：

#include<bits/stdc++.h>
using namespace std;
bool isprime(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;cin>>n;bool flag=false;
    if(isprime(n))cout<<"1
"<<n<<endl;//如果本身是素数，直接输出即可
    else{
        for(int i=3;i<=n;i+=2){//从3开始按奇数来枚举
            if(isprime(i)){
                int tmp=n-i;
                if(isprime(tmp)){cout<<"2
"<<i<<' '<<tmp<<endl;break;}
                for(int j=3;j<=n;j+=2)//从3开始按奇数来枚举
                    if(isprime(j) && isprime(tmp-j)){cout<<"3
"<<i<<' '<<j<<' '<<(tmp-j)<<endl;flag=true;break;}
                if(flag)break;
            }
        }
    }
    return 0;
}