Codeforces 735D - Taxes 打表

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 + ... + 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 打表发现只有1，2,3 这几种情况

#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <queue> #include <stack> #include <vector> #include <list> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) const int inf_int = 2e9; const long long inf_ll = 2e18; #define inf_add 0x3f3f3f3f #define mod 1000000007 #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define pi acos(-1.0) #define pii pair<int,int> #define Lson L, mid, rt<<1 #define Rson mid+1, R, rt<<1|1 const int maxn=5e2+10; using namespace std; typedef long long ll; typedef unsigned long long ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1; while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-') fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48, rx=getchar();return ra*fh;} //#pragma comment(linker, "/STACK:102400000,102400000") ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} bool check(long long x){ for(int i = 2; i*i <= x; ++i)if(x%i == 0)return false; return true; } int main(){ long long n; scanf("%I64d", &n); if(check(n))puts("1"); else if(n%2 == 0 || check(n-2))puts("2"); else puts("3"); }