Codeforces Round #174 (Div. 2) Problem A

A. Cows and Primitive Roots

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The cows have just learned what a primitive root is! Given a prime p, a primitive root  is an integer x (1 ≤ x < p) such that none of integers x - 1, x² - 1, ..., x^p - 2 - 1 are divisible by p, but x^p - 1 - 1 is.

Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime p, help the cows find the number of primitive roots .

Input

The input contains a single line containing an integer p (2 ≤ p < 2000). It is guaranteed that p is a prime.

Output

Output on a single line the number of primitive roots .

Sample test(s)

input

3

output

1

input

5

output

2

Note

The only primitive root  is 2.

The primitive roots  are 2 and 3.

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <iomanip>using namespace std;int main(){int p;cin>>p;int ans=0;for(int x = 1; x < p; x++) //遍历x {int tmp=1;for(int j=1; j<p; j++) //x的j次方  {tmp*=x;tmp%=p;   //防止数据溢出，求余操作分歩 if(tmp==1)//即满足（x^tmp - 1）% p == 0 {if(j==p-1) // 即符合题意 （x^(p-1) - 1）% p == 0 ans++;break;}}}cout<<ans<<endl;return 0;}