Codeforces Round #382 (Div. 2) 735D - Taxes

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 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

题意就是给你一个数N，然后把这个数分成任意份大小不同的数，把这个数看成一份也可以，还有就是每份都要大于1，你要怎么分才能使每份数除自己外的最大因数加起来最小。要每个数除自己外的最大因数最小那就会想到素数，因此要尽量使这个给出的数分成几份素数，这其中可以用到已被证明的哥德巴赫猜想：大于等于4的偶都可以由两个相加得到。所以除2以外的偶数都只需要输出2即可，2的情况特判一下；如果是奇数，先判断是否为素数，在判断这个数N-2是否为素数，如果都不是的话我们还可用上面的哥德巴赫猜想推导出大于等于7的奇数都可以由3个素数相加得到，输出3就行！

博主就是在比赛时卡在了没有判断N-2为素数的情况，当时脑子秀逗了。代码很短，就是脑洞不够，好题赞一个！

代码：

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