The cows have just learned what a primitive root is! Given a prime *p*, a primitive root <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f203bd9f1553a2d666452cec7d6f1f4f02a817fb.png" style="max-width: 100.0%;max-height: 100.0%;"/> is an integer *x* (1 ≤ *x* &lt; *p*) such that none of integers *x* - 1, *x*<sup class="upper-index">2</sup> - 1, ..., *x*<sup class="upper-index">*p* - 2</sup> - 1 are divisible by *p*, but *x*<sup class="upper-index">*p* - 1</sup> - 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 <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f203bd9f1553a2d666452cec7d6f1f4f02a817fb.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
The input contains a single line containing an integer *p* (2 ≤ *p* &lt; 2000). It is guaranteed that *p* is a prime.
#include<bits/stdc++.h>
using namespace std ;
int main()
{
    int cnt =0 ; 
    int n ; cin>>n ;
    for(int i = n -1; i >=1 ; i--)
    {
        bool res = 1 ;
        int cur = i ; 
        for(int j =1 ; j<=n-1 && res;j++,cur =(i*cur)%n )
            if((n %cur != 0 && j == n-1) ||
             (n % cur ==0 && j != n-1))
                res &=0;

        if(res )
            cnt++;
        
    }
    cout<<cnt<<endl;
}