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.

The first line of the input contains a single integer n (2 ≤ n ≤ 2·10⁹) — the total year income of mr. Funt.

Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.

4

2

27

#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define mod 1000000007

#define esp 0.00000000001

const int N=1e5+,M=1e6+,inf=1e9;

const ll INF=1e18+;

int prime(int n)

{

    if(n<=)

    return ;

    if(n==)

    return ;

    if(n%==)

    return ;

    int k, upperBound=n/;

    for(k=; k<=upperBound; k+=)

    {

        upperBound=n/k;

        if(n%k==)

            return ;

    }

    return ;

}

int main()

{

    int x;

    scanf("%d",&x);

    if(prime(x))

        return puts("");

    if(x%==)

        return puts("");

    if(prime(x-))

        return puts("");

    puts("");

    return ;

}