任一大于5的整数都可写成三个质数之和。
(n>5：当n为偶数，n=2+(n-2)，n-2也是偶数，可以分解为两个质数的和；
当n为奇数，n=3+(n-3)，n-3也是偶数，可以分解为两个质数的和)。

题目描述
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 + ... + nk = 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 ni ≥ 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.

题目大意：
给你一个n（2≤n≤2e9） 代表一个人的收入。

他需要交税，规则：交税金额为n的最大公约数（本身不算）

他想通过把钱分成几份，然后分别交税，达到交税最少。

输入
The first line of the input contains a single integer n (2 ≤ n ≤ 2·10^9) — the total year income of mr. Funt.

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

代码

#include <bits/stdc++.h>
using namespace std;
long long n;
int sushu(long long x){
    for(long long i=2;i<=sqrt(x);i++){
        if(!(x%i))return 0;
    }
    return 1;
}
int main(){
    cin>>n;
    if(n%2==0&&n!=2)cout<<2<<endl;
    else if(sushu(n))cout<<1<<endl;
    else if(sushu(n-2))cout<<2<<endl;
    else cout<<3<<endl;
    return 0;
}