Codeforces#382(Div. 2) D.Taxes【哥德巴赫猜想】_outputstandard output mr. funt now lives in a coun-优快云博客

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介绍了一种税收最小化的方法，通过将总收入拆分成多个部分并分别计算税款，以减少总税负。对于不同类型的收入（偶数或奇数），给出了具体的拆分策略。

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Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.

As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n₁ + n₂ + ... + n_k = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition n_i ≥ 2 should hold for all i from 1 to k.

Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 2·10⁹) — the total year income of mr. Funt.

Output

Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.

Examples

Input

Output

Input

Output

题目大意：

给你一个数N，然你找几个数，使得其加和值为N。使得使用的数字量最少。

思路：

1、我们分类讨论，首先讨论偶数的情况：

根据强哥德巴赫猜想，我们能够知道，任何大于2的偶数能够拆成两个素数相加得到，那么明显，如果N==2，那么其本身也是素数，那么输出1，否则输出2.

2、接下来讨论奇数的情况：

①如果这个数本身就是素数，那么很肯定，输出1即可。

②如果这个数是两个数相加得到的，然而这个数是奇数，很明显只能通过奇数+偶数==奇数得来的，然而偶数中素数只有2，那么判定一下n-2是不是素数即可，如果是，那么输出2，否则输出3。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
#define ll __int64
int isprime(ll a)
{
    for(int i=2;i<=sqrt(a);i++)
    {
        ll tmp=i;
        if(a%tmp==0)return 0;
    }
    return 1;
}
int main()
{
    ll n;
    while(~scanf("%I64d",&n))
    {
        if(isprime(n)==1)
        {
            printf("1\n");
            continue;
        }
        if(n%2==1)
        {
            if(isprime(n-2)==1)printf("2\n");
            else printf("3\n");
        }
        if(n%2==0)
        {
            if(n==2)
            {
                printf("1\n");
            }
            else printf("2\n");
        }
    }
}