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 sqrt
from array import array
from sys import stdin
input =  stdin.readline
n = int(input())
ans = []
if n == 3:
	ans.append(3)
	print('{}'.format(len(ans)) + '\n' + ''.join(str(x) + ' ' for x in ans))
	exit(0)
N = n
magic_value = 300
primes = array('L')
primes.append(2)
lim = int(sqrt(N)) + 1

for number in range(3, lim, 2):
	isPrime = True
	for prime in primes:
		if number % prime == 0:
			isPrime = False
			break
	if isPrime:
		primes.append(number)
def checkPrime(x):
	lim = int(sqrt(x) + 1)
	isPrime = True
	for prime in primes:
		if x % prime == 0:
			isPrime = False
			break
		if prime > lim:
			break
	return isPrime
for sub in range(0, magic_value, 2):
	if checkPrime(n - sub):
		ans.append(n - sub)
		n = sub
		break
if n == 0:
	print('{}'.format(len(ans)) + '\n' + ''.join(str(x) + ' ' for x in ans))
	exit(0)
for prime in primes:
	if n == prime:
		ans.append(n)
		break
	if n - prime in primes:
		ans.append(prime)
		ans.append(n - prime)
		break
	if prime > n:
		break
print('{}'.format(len(ans)) + '\n' + ''.join(str(x) + ' ' for x in ans))