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+10,M=1e6+10,inf=1e9;
const ll INF=1e18+10;
int prime(int n)
{if(n<=1)return 0;if(n==2)return 1;if(n%2==0)return 0;int k, upperBound=n/2;for(k=3; k<=upperBound; k+=2){upperBound=n/k;if(n%k==0)return 0;}return 1;
}
int main()
{int x;scanf("%d",&x);if(prime(x))return puts("1");if(x%2==0)return puts("2");if(prime(x-2))return puts("2");puts("3");return 0;
}