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;
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC target("avx")
const long long INF = numeric_limits<long long>::max() / 2;
const long double PI = 3.1415926535898;
const long long MOD = 1e9 + 7;
const long long LIM = 1e5 + 5;
long long fpow(long long x, long long y) {
  long long temp;
  if (y == 0) return 1;
  temp = fpow(x, y / 2);
  if (y % 2 == 0)
    return (temp * temp) % MOD;
  else
    return (x * ((temp * temp) % MOD)) % MOD;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long modInverse(long long n, long long p) { return fpow(n, p - 2) % p; }
map<long long, long long> primeFactors(long long n) {
  map<long long, long long> mp;
  while (n % 2 == 0) {
    if (mp.count(2) == 1)
      mp[2]++;
    else
      mp.insert({2, 1});
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      if (mp.count(i) == 1)
        mp[i]++;
      else
        mp.insert({i, 1});
      n = n / i;
    }
  }
  if (n > 2) {
    if (mp.count(n) == 1)
      mp[n]++;
    else
      mp.insert({n, 1});
  }
  return mp;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> ans;
  if (isPrime(n))
    ans.push_back(n);
  else if (isPrime(n - 2)) {
    ans.push_back(2);
    ans.push_back(n - 2);
  } else {
    ans.push_back(3);
    n = n - 3;
    for (long long i = 0; i < n; i++) {
      if (isPrime(i) && isPrime(n - i)) {
        ans.push_back(i);
        ans.push_back(n - i);
        break;
      }
    }
  }
  cout << ans.size() << "\n";
  for (long long x : ans) cout << x << " ";
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  { solve(); }
  return 0;
}