Codeforces Round #388 (Div. 2) A

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
题意：将数字分解成素数之和形式，要求素数个数尽量多
解法：作死的填2和3就成

#include<bits/stdc++.h>
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
using namespace std;
#define debug(x) cout << #x" = " << x<<endl;
int main(){
    int n;
    cin>>n;
    cout<<n/2<<endl;
    if(n%2){
        for(int i=1;i<n/2;i++){
            cout<<"2 ";
        }
        cout<<"3"<<endl;
    }else{
        for(int i=1;i<=n/2;i++){
            cout<<"2 ";
        }
    }
    return 0;
}