Codeforces Round #388 (Div. 2) 749A Bachgold Problem 【素数】

本文链接：https://blog.youkuaiyun.com/qq_37405320/article/details/75725513

本文介绍了Bachgold问题，即给定一个正整数n，将其表示为尽可能多的质数之和。文章提供了解决这一问题的具体算法思路，并附带了一个简单的C++实现示例。

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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 than1.

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

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 ton. You can print them in any order. If there are several optimal solution, print any of them.

Examples

Input

Output

2
2 3

Input

Output

3
2 2 2

思路：给你一个数n，问你最多能把n分成几个素数的和？输出素数的数量和这些素数。

思路：如果n是偶数的话。那就可以分成n/2个2

如果n是奇数的话。那就可以分成(（n-1）/2 -1)个2和一个3

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define M(a)  memset(a,0,sizeof(a))
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==2)
        {
            printf("1\n");
            printf("2\n");
            continue;
        }
        if(n%2==0)
        {
            printf("%d\n",n/2);
            for(int i=0;i<n/2;i++)
            {
                if(i==0)
                {
                    printf("2");
                }
                else
                {
                    printf(" 2");
                }
            }
            printf("\n");
        }
        else
        {
            printf("%d\n",n/2);
            for(int i=0;i<n/2-1;i++)
            {
                printf("2 ");
            }
            printf("3\n");
        }
    }
    return 0;
}