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

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1. Codeforces Round #382 (Div. 2) — D. 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 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



#include
typedef long long ll;
#define maxn 100008
#define cle(n) memset(n,0,sizeof(n))
using namespace std;

int n,m,k;

int prime(int n)
{
    for(int i=2;i*i<=n;i++)
    {
        if(n%i==0)
            return 0;
    }
    return 1;
}
int main()
{
   cin>>n;
   if(prime(n))
    return cout<<1<







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