Codeforces Round #382 (Div. 2) D Taxes（数论）

题意：你有n块钱，有两个方式给钱，一种是给n的最大因子，另外一种是把n拆成大于等于2个数，给这些数的最大因子和，求最小的给钱数

思路：由哥德巴赫猜想可知，任意一个>2的偶数可以写成两个质数的和，任意一个>7的奇数可以写成3个质数的和，那么把2，3，4，5，6，7的情况写出来，然后判是否素数，然后分情况讨论即可，有一个特别的是如果是>7的奇数要先判n-2是否为素数，如果是的话那么就是2，否则就是3

#include<bits/stdc++.h>
using namespace std;
#define LL long long
bool check(LL n)
{
	for(int i = 2;i<=sqrt(n);i++)
		if(n%i==0)return false;
	return true;
}
int main()
{
   LL n;
   scanf("%lld",&n);
   if(n==2)printf("1\n");
   else if(n==3)printf("1\n");
   else if(n==4)printf("2\n");
   else if(n==5)printf("1\n");
   else if(n==6)printf("2\n");
   else if(n==7)printf("1\n");
   else if(check(n))printf("1\n");
   else if(n%2==0)printf("2\n");
   else if(check(n-2)) printf("2\n");
   else printf("3\n");
}

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