Codeforces Round #382 (Div. 2) -- D. Taxes

D. 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 + ... + 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.

Examples

input

4

output

2

input

27

output

3

#include<bits/stdc++.h>typedef long long ll;#define maxn 100008#define cle(n) memset(n,0,sizeof(n))using namespace std;int n,m,k;int prime(int n){    for(int i=2;i*i<=n;i++)    {        if(n%i==0)            return 0;    }    return 1;}int main(){   cin>>n;   if(prime(n))    return cout<<1<<endl,0;   if(n%2==0)    return  cout<<2<<endl,0;   else   {       if(prime(n-2))        return cout<<2<<endl,0;       else cout<<3<<endl,0;   }    return 0;}