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

题目大意：给一个正整数N，判断它最多是多少个素数的和？并输出这些素数。

思路：因为要个数最多，最简单的想法就是，只用2和3。偶数就拆成（n/2）个2，奇数就拆成（（n-3）/2）个2和1个3。

code:

#include<iostream>
#include<stdio.h>
using namespace std;

int main()
{
int n;
scanf("%d",&n);
if(n % 2 == 0)
{
n /= 2;
printf("%d\n",n);
for(int i=1; i<n; i++)
printf("%d ",2);
printf("%d\n",2);
}
else
{
n -= 3;
n /= 2;
printf("%d\n",n+1);
for(int i=1;i<=n;i++)
printf("%d ",2);
printf("%d\n",3);
}
return 0;
}