*CF2.D（哥德巴赫猜想）

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 + ... + 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：
有n元钱，n>1，要缴纳n的最大质因数的税，但可以把n分成若干份缴纳每一份最大质因数的和，其中不能有1。
代码：

//哥德巴赫猜想：任意一个大于2的偶数可以拆成两个素数的和，任意一个大于7的奇数可以拆成三个素数的和。
//本题答案是1.判断n是否是素数。是2.判断是否是偶数；猜想的逆就是除了2以外任意两个素数的和都是偶数所以还要判断n-2是不是素数。否则就是三3。
#include<bitsstdc++.h>
using namespace std;
int sus(int n)
{
    int tem=sqrt(n);
    for(int i=2;i<=tem;i++)
    if(n%i==0) return 0;
    return 1;
}
int main()
{
    int n;
    scanf("%d",&n);
    if(sus(n))
    printf("1
");
    else if(n%2==0||sus(n-2))
    printf("2
");
    else printf("3
");
    return 0;
}