Problem: Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.

Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k. 

Input

The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).

Output

The first line of the output contains a single integer k — maximum possible number of primes in representation.

The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.

Examples

Input

5


Output

2
2 3


Input

6


Output

3
2 2 2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool prime[1000001];
void sieve(int n) {
  int x = sqrt(n);
  prime[0] = prime[1] = 1;
  for (int i = 4; i <= n; i += 2) prime[i] = 1;
  for (int i = 3; i <= x; i += 2) {
    if (prime[i] == 0) {
      for (int j = i + i; j <= n; j += i) prime[j] = 1;
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  sieve(1000000);
  long long int t, a, b, c, d, i;
  cin >> a;
  if (a % 2 == 0) {
    b = a / 2;
    cout << b << endl;
    for (int i = 0; i < b; i++) {
      cout << 2 << " ";
    }
    cout << endl;
  } else {
    c = (a / 2) - 1;
    cout << a / 2 << endl;
    for (int i = 0; i < c; i++) {
      cout << 2 << " ";
    }
    cout << 3 << endl;
  }
}