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 CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
template <class T>
inline void gmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline void gmin(T &a, T b) {
  if (b < a) a = b;
}
using namespace std;
const int N = 1e5 + 10, M = 2e6 + 10, Z = 1e9 + 7, maxint = 2147483647,
          ms1 = 16843009, ms31 = 522133279, ms63 = 1061109567,
          ms127 = 2139062143;
const double PI = acos(-1.0), eps = 1e-8;
void fre() {
  freopen("/Users/luras/Desktop/in.txt", "r", stdin);
  freopen("/Users/luras/Desktop/out.txt", "w", stdout);
}
const int INF = 1e9;
int casenum, casei;
const int S = 8;
long long mult_mod(long long a, long long b, long long c) {
  a %= c;
  b %= c;
  long long ret = 0;
  long long tmp = a;
  while (b) {
    if (b & 1) {
      ret += tmp;
      if (ret > c) ret -= c;
    }
    tmp <<= 1;
    if (tmp > c) tmp -= c;
    b >>= 1;
  }
  return ret;
}
long long pow_mod(long long a, long long n, long long mod) {
  long long ret = 1;
  long long temp = a % mod;
  while (n) {
    if (n & 1) ret = mult_mod(ret, temp, mod);
    temp = mult_mod(temp, temp, mod);
    n >>= 1;
  }
  return ret;
}
bool check(long long a, long long n, long long x, long long t) {
  long long ret = pow_mod(a, x, n);
  long long last = ret;
  for (int i = 1; i <= t; i++) {
    ret = mult_mod(ret, ret, n);
    if (ret == 1 && last != 1 && last != n - 1) return true;
    last = ret;
  }
  if (ret != 1)
    return true;
  else
    return false;
}
bool is_prime(long long n) {
  if (n < 2) return false;
  if (n == 2) return true;
  if ((n & 1) == 0) return false;
  long long x = n - 1;
  long long t = 0;
  while ((x & 1) == 0) {
    x >>= 1;
    t++;
  }
  srand(time(NULL));
  for (int i = 0; i < S; i++) {
    long long a = rand() % (n - 1) + 1;
    if (check(a, n, x, t)) return false;
  }
  return true;
}
int n;
int main() {
  while (~scanf("%d", &n)) {
    if (n == 3)
      printf("1\n3\n");
    else if (n == 4)
      printf("2\n2 2\n");
    else if (n == 5)
      printf("2\n2 3\n");
    else if (n == 6)
      printf("2\n3 3\n");
    else if (n == 7)
      printf("2\n2 5\n");
    else if (n == 8)
      printf("3\n2 3 3\n");
    else {
      printf("3\n");
      n -= 3;
      printf("3 ");
      for (int i = 3; i <= n; i += 2) {
        if (is_prime(i) && is_prime(n - i)) {
          printf("%d %d\n", i, n - i);
          break;
        }
      }
    }
  }
  return 0;
}