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.
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.
读题易想到所有素数都只算1的税，所以联想到分解成素数（本身素数那就是1）。然后根据哥德巴赫猜想，所有偶数可以分解为两个素数的和，所以除了2以外的所有偶数答案都是2，而对于非素数奇数就是看他能不能分解2+素数的形式，否则就是3
#include <iostream>
#include <iomanip>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
LL n;
LL res = 0;
bool isprime(LL n){
if (n == 2){
return true;
}
for (int i=2;i<=sqrt(n);i++){
if (n % i == 0){
return false;
}
}
return true;
}
int main(){
// freopen("test.in","r",stdin);
cin >> n;
if (isprime(n)){
cout << 1; return 0;
}
if (n % 2 == 0){
cout << 2;
}
else {
if (isprime(n-2)){
cout << 2;
}
else
cout << 3;
}
return 0;
}