Problem: 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. <image>



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.

Examples

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.

Solution (in PYTHON3):

Solution:
from math import *

def estpremier(n):
    """estpremier(n): dit si un nombre est premier (renvoie True ou False)"""
    if n<7:
        if n in (2,3,5):
            return True
        else:
            return False
    # si n est pair et >2 (=2: cas traité ci-dessus), il ne peut pas être premier
    if n & 1 == 0:
        return False
    # autres cas
    k=3
    r=int(sqrt((n)))+1
    while k<=r:
        if n % k == 0:
            return False
        k+=2
    return True

n=int(input())
if estpremier(n):
	print(1)
	print(n)
else:
	bb=True
	m=(n+1)//2
	z=1000
	i=max(2,m-z)
	while bb and i<m+1:
		if estpremier(i):
			j=n-i
			if estpremier(j):
				bb=False
				print(2)
				print(i,j)
		i+=1
	
	if bb:
		m=(n+1)//3
		z=1000
		i=max(2,m-z)
		while bb and i<m+1:
			if estpremier(i):
				j=i
				while bb and j<m+z:
					if estpremier(j):
						k=n-i-j
						if estpremier(k):
							bb=False
							print(3)
							print(i,j,k)
					j+=1
			i+=1