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>
const double pi = 3.14159265359;
const int MOD = 1000000007;
const int dr[4] = {-1, 1, 0, 0};
const int dc[4] = {0, 0, -1, 1};
double sinA(double dig) { return sin(dig * pi / 180); }
double cosA(double dig) { return cos(dig * pi / 180); }
double tanA(double dig) {
  int we = dig;
  if (we == dig) {
    if (we % 90 == 0) {
      if ((we / 90) % 2 == 1)
        std::cout << "ERROR\n";
      else
        return 0;
    }
  }
  return tan(dig * pi / 180);
}
double secA(double dig) {
  int qw = cos(dig * pi / 180);
  if (qw == 0)
    std::cout << "ERROR\n";
  else
    return (1 / qw);
}
double cosecA(double dig) {
  int qw = sin(dig * pi / 180);
  if (qw == 0)
    std::cout << "ERROR\n";
  else
    return (1 / qw);
}
double cotA(double dig) {
  int qw = tan(dig * pi / 180);
  if (qw == 0)
    std::cout << "ERROR\n";
  else
    return (1 / qw);
}
bool isPrime(long long num) {
  if (num < 0) num *= -1;
  if (num == 1 || num == 0)
    return false;
  else if (num == 2)
    return true;
  else {
    if (num % 2 == 0) {
      return false;
    } else
      for (int x = 3; x <= sqrt(num); x += 2) {
        if (num % x == 0) {
          return false;
        }
      }
    return true;
  }
}
long long gcd(long long A, long long B) {
  if (B == 0) return A;
  return gcd(B, A % B);
}
long long lcm(long long f, long long l) { return (f / gcd(f, l) * l); }
long long fact(long long number) {
  if (number < 0) number *= -1;
  if (number == 0) return 1;
  if (number == 1)
    return number;
  else
    return number * fact(number - 1);
}
long long fib(long long n) {
  auto sqrt_5 = std::sqrt(5);
  if (n == 0) return 0;
  if (n == 1) return 1;
  return static_cast<long long>(
      (std::pow(1 + sqrt_5, n) - std::pow(1 - sqrt_5, n)) /
      (std::pow(2, n) * sqrt_5));
}
long long cdiv(long long a, long long b) {
  return a / b + ((a ^ b) > 0 && a % b);
}
long long fdiv(long long a, long long b) {
  return a / b - ((a ^ b) < 0 && a % b);
}
int power(int x, int n) {
  if (n == 0) return 1;
  int temp = power(x, n / 2);
  temp *= temp;
  if (n % 2) temp *= x;
  return temp;
}
long long powerMOD(long long x, long long n, long long m) {
  if (n == 0) return (1);
  long long temp = powerMOD(x, n / 2, m);
  temp = ((temp % m) * (temp % m)) % m;
  if (n % 2) temp = ((temp % m) * (x % m)) % m;
  return temp;
}
int ETF(int n) {
  int ans = 1;
  for (int x = 2; x * x <= n; x++) {
    if (n % x == 0) {
      int count = 0;
      while (n % x == 0) {
        count++;
        n /= x;
      }
      ans *= (power(x, count - 1) * (x - 1));
    }
  }
  if (n != 1) ans *= (n - 1);
  return ans;
}
long long SUM(long long number) {
  long long ans = 0;
  while (number) {
    ans += number % 10;
    number /= 10;
  }
  return ans;
}
std::map<long long, long long> f;
long long fibb(long long n) {
  if (f.count(n)) return f[n];
  if (n == 0) return 0;
  if (n == 1 || n == 2) return 1;
  if (n % 2 == 0) {
    long long k = n / 2;
    long long ret1 = fibb(k - 1), ret2 = fibb(k);
    return f[n] = ((((2ll * ret1) % MOD + ret2) % MOD) * ret2) % MOD;
  } else {
    long long k = (n + 1) / 2;
    long long ret1 = fibb(k - 1), ret2 = fibb(k);
    return f[n] = ((ret1 * ret1) % MOD + (ret2 * ret2) % MOD) % MOD;
  }
}
void generateNthrow(int N) {
  int prev = 1;
  std::cout << prev;
  for (int i = 1; i <= N; i++) {
    int curr = (prev * (N - i + 1)) / i;
    std::cout << ", " << curr;
    prev = curr;
  }
}
int logNM(int num1, int num2) {
  int counter = 0;
  while (num2) {
    num2 /= num1;
    counter++;
  }
  return counter + 1;
}
int pr(int number) {
  for (int x = 2; x * x <= number; x++) {
    if (number % x == 0) return 0;
  }
  return 1;
}
void run() {
  int num;
  std::cin >> num;
  if (pr(num))
    std::cout << "1\n" << num;
  else {
    std::cout << "3\n3 ";
    num -= 3;
    for (int x = 3; x < num; x += 2) {
      if (pr(x) && pr(num - x)) {
        std::cout << x << " " << num - x;
        return;
      }
    }
  }
}
int main() { run(); }