Bachgold Problem CodeForces - 749A

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.

Example

Input

5

Output

2
2 3

Input

6

Output

3

2 2 2
题意：给定一个数n，求能被最大数目的质数加起来等于n，求最多的数目；
分析：最小的 质数除了1,就是2,3，所以用他俩来加，无疑是数目最多的。此题要注意细节，比如空格什么的

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <queue>
#include <string.h>
using namespace std;
int main()
{
int n;
while(cin>>n)
{
int num=0;
if(n%2==0)
{
num=n/2;
printf("%d\n",num);
for(int i=0; i<num; i++)
{
if(i!=0)
printf(" ");
printf("2");
}
printf("\n");
}
else
{
num=(n-3)/2+1;
printf("%d\n",num);
if(n==3)
printf("3\n");
else
{
for(int i=0; i<num-1; i++)
{
printf("2 ");
}
printf("3\n");
}

}
}
}