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

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 n

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

題意：某人要交稅，交的稅錢是收入n的最大因子（!=n，若該數是質數則是1），但是這人作弊，把錢拆成幾份，使交稅最少，輸出稅錢。

解題思路：哥德巴赫猜想求解。

1.如果數字本身就是素數，直接輸出1。

2.如果是2的話，直接輸出1 （因為2也是素數）。

3.偶數的話，直接輸出2（因為任何一個偶數都可以拆分成兩個素數）。

4.如果是奇數的話，需要看看n-2是不是素數，若n-2是素數的話，就直接輸出2（拆分成2  n-2）

5.如果是奇數的話，且n-2不是素數，那麼就直接輸出3 ，所有的情況都已經考慮完畢。

ac程式碼：

#include<bits/stdc++.h>
#include<iostream>
using namespace std;
typedef long long ll;
ll n;
int ok(int x)
{
	for(int i=2;i<=sqrt(x);i++)
	{
		if(x%i==0)
			return 0;
	}
	return 1;
}
int main()
{
	scanf("%lld",&n);
	if(n==2)
		printf("1\n");
	else if(n%2==0)
		printf("2\n");
	else if(ok(n))
		printf("1\n");
	else if(ok(n-2))
		printf("2\n");
	else
		printf("3\n");
	return 0;
}