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>
using namespace std;
float PHI = 1.6180339;
const float PI_F = 3.14159265358979f;
long long modpow(long long x, long long n, long long m) {
  if (n == 0) return 1 % m;
  long long u = modpow(x, n / 2, m);
  u = (u * u) % m;
  if (n % 2 == 1) u = (u * x) % m;
  return u;
}
long long add(long long x, long long y, long long m) {
  x += y;
  if (x >= m) return x - m;
  return x;
}
long long sub(long long x, long long y) {
  x -= y;
  if (x < 0) return x + 1000000007;
  return x;
}
long long inv(long long a, long long m = 998244353) {
  return modpow(a, m - 2, m);
}
long long mul(long long a, long long b, long long m) {
  return (long long)a * b % m;
}
long long nCrModp(long long n, long long r, long long p) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, r); j > 0; j--) C[j] = (C[j] + C[j - 1]) % p;
  }
  return C[r];
}
bool sortinrev(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.first - a.second) > (b.first - b.second);
}
vector<long long> z_function(string &s) {
  long long n = s.size();
  vector<long long> z(n);
  for (long long i = 1, l = 0, r = 0; i < n; i++) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) z[i]++;
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z;
}
bool isprime(long long n) {
  if (n <= 1) return false;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
void SieveOfEratosthenes(long long n, bool isPrime[]) {
  isPrime[0] = isPrime[1] = false;
  for (long long i = 2; i <= n; i++) isPrime[i] = true;
  for (long long p = 2; p * p <= n; p++) {
    if (isPrime[p] == true) {
      for (long long i = p * p; i <= n; i += p) isPrime[i] = false;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  cout << fixed << setprecision(16);
  long long n;
  cin >> n;
  if (isprime(n)) {
    cout << "1"
         << "\n";
    cout << n << "\n";
    return 0;
  }
  long long maxn = 1e3;
  long long x = n;
  while (true) {
    if (isprime(x)) {
      break;
    }
    --x;
  }
  long long left = n - x;
  if (isprime(left)) {
    cout << "2"
         << "\n";
    cout << x << " " << left << "\n";
    return 0;
  }
  bool isPrime[maxn + 10];
  SieveOfEratosthenes(left, isPrime);
  long long f = 0, g = 0;
  for (long long i = 0; i < left; i++) {
    if (isPrime[i] && isPrime[left - i]) {
      f = i;
      g = left - i;
      break;
    }
  }
  cout << "3"
       << "\n";
  cout << x << " " << f << " " << g << "\n";
  return 0;
}