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 5and 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 ≥ 2should 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
4
Output
2
Input
27
Output
3
题意：某人要交税，交的税钱是收入n的最大因子（!=n，若该数是质数则是1），但是这人作弊，把钱拆成几份，使交税最少，输出税钱。
解题思路：哥德巴赫猜想求解。
1.如果数字本身就是素数，直接输出1。
2.如果是2的话，直接输出1 （因为2也是素数）。
3.偶数的话，直接输出2（因为任何一个偶数都可以拆分成两个素数）。
4.如果是奇数的话，需要看看n-2是不是素数，若n-2是素数的话，就直接输出2（拆分成2 n-2）
5.如果是奇数的话，且n-2不是素数，那么就直接输出3 ，所有的情况都已经考虑完毕。
ac代码：
#include<bits/stdc++.h>
#include<iostream>
using namespace std;
typedef long long ll;
ll n;
int ok(int x)
{
for(int i=2;i<=sqrt(x);i++)
{
if(x%i==0)
return 0;
}
return 1;
}
int main()
{
scanf("%lld",&n);
if(n==2)
printf("1\n");
else if(n%2==0)
printf("2\n");
else if(ok(n))
printf("1\n");
else if(ok(n-2))
printf("2\n");
else
printf("3\n");
return 0;
}