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 PYTHON):

Solution:
import bisect

def index(a, x):
    'Locate the leftmost value exactly equal to x'
    i = bisect.bisect_left(a, x)
    if i != len(a) and a[i] == x:
        return i
    raise ValueError

results = {
    3: "1\n3",
    5: "1\n5",
    7: "1\n7",
    9: "3\n3 3 3",
    11: "1\n11",
    13: "1\n13",
    15: "3\n5 5 5",
    17: "1\n17",
    19: "1\n19",
    21: "3\n7 7 7",
    23: "1\n23",
    25: "3\n3 3 19",
    27: "3\n5 11 11",
    29: "1\n29"
}

def sieve(n):
    "Return all primes <= n."
    np1 = n + 1
    s = list(range(np1)) # leave off `list()` in Python 2
    s[1] = 0
    sqrtn = int(round(n**0.5))
    for i in range(2, sqrtn + 1): # use `xrange()` in Python 2
        if s[i]:
            # next line:  use `xrange()` in Python 2
            s[i*i: np1: i] = [0] * len(range(i*i, np1, i))
    return filter(None, s)



def is_prime(n):
    if n <= 1:
        return False
    elif n <= 3:
        return True
    elif n % 2 == 0 or n % 3 == 0:
        return False
    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i = i + 6

    return True

def closest_prime(number):
    while number > 0:
        if is_prime(number):
            return number
        number -= 1

    return None


primes = sieve(1000)
totals = []
elems = []

for prime1 in primes:
    totals.append((prime1, prime1, None))
    for prime2 in primes:
        totals.append((prime1 + prime2, prime1, prime2))

totals.sort()
totals_search = [elem[0] for elem in totals]

n = int(raw_input())
if n <= 29:
    print results[n]
else:
    prime1 = closest_prime(n - 28)
    to_find = n - prime1
    i = index(totals_search, to_find)
    _, prime2, prime3 = totals[i]
    if prime3 is not None:
        print 3
        print prime1, prime2, prime3
    else:
        print 2
        print prime1, prime2, prime3