CodeForces - 735D Taxes （哥德巴赫猜想）

Taxes

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

4

Output

2

Input

27

Output

3
题意：在一个国家，如果一个人的收入是n，那个这个人所交的税就是n的最大因子（不包括n）
现在你将n拆成多个数的和 规则如上 问你所交的税最少是多少

哥德巴赫猜想：任一大于2的偶数都可写成两个质数之和
所以我们分三种情况
1》这个数本身就是质数  答案为1
2》这是数是一个偶数  答案为2
3》这个数是一个奇数n  n=奇数+2（n>=5) 假如n-2是一个质数 答案为2
假如n-2不为一个质数  我们就可以把这个数化为 n=(n-3)+3  n-3是一个偶数  答案为3

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string.h>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<cmath>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const double PI=acos(-1.0);
const double eps=0.0000000001;
const int INF=1e9;
const int N=100000+100;
int prime[N];
int check(int x){
    for(int i=2;i*i<=x;i++){
        if(x%i==0)return 0;
    }
    return 1;
}
int main(){
    ll n;
    while(scanf("%I64d",&n)!=EOF){
       if(check(n)==1){
            cout<<1<<endl;
            continue;
       }
       if(n%2==0){
            cout<<2<<endl;
            continue;
       }
       else{
            if(check(n-2)==1)cout<<2<<endl;
            else{
                cout<<3<<endl;
            }
       }
    }
}