cf D. Dima and Lisa （三素数定理_素数打表+判定）

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

35 11 11

Note

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

两个素数间差距在800以内（题解是300以内），n-p之后的范围就在2-300之间，且为偶数，根据哥德巴赫猜想，任意大的偶数都可以分解两个质数和。那么通过暴力就能求出余下的了！！！！

#include <bits/stdc++.h>using namespace std;const int maxn=1e6+100;int prime[maxn];bool is_prime(int n){for(int i=2;i*i<=n;i++) {if(n%i==0) return false;}return n!=1;}void init_prime(){int i,j,k;prime[0]=prime[1]=1;for(i=2;i<=maxn;i++) {if(!prime[i])for(j=2*i;j<=maxn;j+=i)prime[j]=1;}}int main(){init_prime();int a,b,c,n,i;cin>>n;if(is_prime(n)) printf("1\n%d\n",n);else if(is_prime(n-2)) printf("2\n2 %d\n",n-2);else {for(i=n;;i--) {if(is_prime(i)) {a=i;break;}}n-=a;for(i=n;;i--) {if(!prime[i] && !prime[n-i]) {b=i;c=n-i;break;}}printf("3\n%d %d %d\n",a,b,c);}return 0;}