Codeforces Round #382 (Div. 1) B. Taxes

B. 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 partsn1 + 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

题意：funt的年收入是n元，交税的费用=n的最大因数（n自身除外），现在他想逃税，把n拆成任意份，但是每一份不能是1，问他最少要交多少钱的税收。

思路：其实这个我楞是没想出来，上网翻了一下题解，原来一个大于2偶数必定能拆成两个素数相加，那么这道题就很简单了，只有几种情况，如果n是2那么答案就是1，如果n是偶数但大于2，那么答案就是2，如果n是奇数且n是素数答案就是1，如果n是奇数但n不是素数n-2是素数的话，答案就是2，如果n是奇数但n不是素数也不是素数的话，答案就是3。下面给代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<queue>
#include<string>
#include<functional>
typedef long long LL;
using namespace std;
#define maxn 105
#define ll l,mid,now<<1
#define rr mid+1,r,now<<1|1
#define lson l1,mid,l2,r2,now<<1
#define rson mid+1,r1,l2,r2,now<<1|1
#define pi acos(-1.0)
#define inf 0x3f3f3f3f
const int mod = 1e9 + 7;
int main()
{
	LL n;
	while (~scanf("%lld", &n))
	{
		if (n == 2){
			printf("1\n");
		}
		else if (n % 2){
			bool jud1 = false, jud2 = false;
			for (int i = 2; i <= sqrt(n); i++){
				if (!(n%i)){
					jud1 = true;
					break;
				}
			}
			n -= 2;
			for (int i = 2; i <= sqrt(n); i++){
				if (!(n%i)){
					jud2 = true;
					break;
				}
			}
			if (!jud1)
				printf("1\n");
			else if (jud1&&jud2)
				printf("3\n");
			else
				printf("2\n");
		}
		else
			printf("2\n");

	}
}