Codeforces Round #324 (Div. 2) D. Dima and Lisa (Goldbach conjecture + violence)

D. Dima and Lisa

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

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

1. 1 ≤ k ≤3
2. P i is a prime

The numbers P I do not necessarily has to be distinct. It is guaranteed, at least one possible solution exists.

Input

The single line contains an odd number n (3≤ n <9).

Output

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

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

Examplesinput

27

Output

3
5 11 11

Note

A Prime is an integer strictly larger than one, which is divisible A-by-one and by itself.

Test instructions is the sum of an odd n split into three or less primes.

According to Goldbach conjecture, any odd number greater than 7 can be written as a sum of three prime numbers. So we randomly find an odd number A as the first prime number, so that the n-a is even, according to the original Goldbach conjecture, any even greater than 2 can be written in the sum of two prime numbers, we can rest assured to find the remaining two primes. The distance between primes is only hundreds of, so you can actually be assured of violence (but I don't know). So I set the Miller-robin algorithm to detect the prime number (as a trial template).

#include <cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>#include<cmath>#include<cstdlib>#include<ctime>using namespacestd;#definell Long LongConstll mod = 1e9 +7;Const intMAXN = 1e5 +Tenll Mod_mul (ll A, ll B, ll N) {ll res=0;  while(b) {if(b&1) Res = (res + a)%N; A= (A + a)%N; b>>=1; }    returnRes;} ll Mod_exp (ll A, ll B, ll N) {ll res=1;  while(b) {if(b&1) res =Mod_mul (Res, a, n); A=Mod_mul (A, A, n); b>>=1; }    returnRes;}BOOLMiller_rabin (ll N) {if(n = =2|| n = =3|| n = =5|| n = =7|| n = = One)return true; if(n = =1|| ! (n%2) || ! (n%3) || ! (n%5) || ! (n%7) || ! (n% One))return false;    ll x, pre, U; intI, j, k =0; U= N-1;  while(! (u&1) ) {k++; U >>=1; } srand ((ll) Time (0));  for(i =0; I < -; ++i) {x= rand ()% (n2) +2; if((x%n) = =0)Continue; X=mod_exp (x, u, n); Pre=x;  for(j =0; J < K; ++j) {x=Mod_mul (x, x, N); if(x = =1&& Pre! =1&& Pre! = N1)return false; Pre=x; }        if(X! =1)return false; }    return true;}intMain () {ll n;  while(~SCANF ("%i64d", &N)) {ll A; if(n = =3|| n = =5) printf ("1\n%i64d\n", N); Else {            if(n = =9) {printf ("2\n2 7\n");return 0;} if(n = =7) {printf ("1\n7\n");return 0;} A= N/3; ifA2==0) a++;  while(1) {a+=2; if(Miller_rabin (a)) Break; } puts ("3");            ll B, C;  for(b =3;; B + =2) {C= N-b-A; //cout << b << "<< C <<" "<< a << Endl;                if(Miller_rabin (b) &&Miller_rabin (c)) {                     Break; }} printf ("%i64d%i64d%i64d\n", A, b, c); }    }}

Codeforces Round #324 (Div. 2) D. Dima and Lisa (Goldbach conjecture + violence)