CodeForces

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 + ... + 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.

Example
Input
4
Output
2
Input
27
Output

3

给出一个数，要尽量把它拆解成素数，而且数量尽量小，据哥德巴赫猜想，一个大于二的偶数一定可以写成两个素数的和， 对于这题

1.n是素数时答案是1；

2.n是偶数    2;

3.n是奇数：

    a.n-2是素数  2；

    b.否则答案是  3；

#include <cstdio>#include <algorithm>using namespace std;bool isPrime(int n){if(n<2)return false;if(n==2||n==3)return true;for(int i=2;i*i<=n;i++)if(n%i==0)return false;return true;}int main(){int n;scanf("%d", &n);if(isPrime(n))printf("1");else if(n%2==0)printf("2");else if(isPrime(n-2))printf("2");else printf("3");return 0;}