code forces 382 D Taxes（数论--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

【分析】给你一个数n，让你分成k个大于1的数，没个数取他们的最大因子，可以是1但不可以是本身，求最小的因子和。显然如果分成k个
质数的和，那么答案就是k.根据哥德巴赫猜想，任何一个偶数都可以写成两个质数的和，所以对于偶数（除了2），答案就是2，对于质数，
答案就是1，对于非质数的奇数，若可写成2+质数，答案就是2，否则3.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 2e9
#define met(a,b) memset(a,b,sizeof a)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long ll;
using namespace std;
const int N = 1e5+;
const int M = 4e5+;
int dp[N][];
int n,sum[N],m=,p,k;
int Tree[N];
bool isprime(ll n){
    if(n<=)return false;
    if(n==)return true;
    else if(n%==)return false;
    for(int i=;(ll)i*i<=n;i+=)if(n%i==)return false;
    return true;
}
int main()
{
    ll q;
    scanf("%lld",&q);
    if(isprime(q))puts("");
    else {
        if(q%==)puts("");
        else {
            if(isprime(q-))puts("");
            else puts("");
        }
    }
    return ;
}