Do you have spent some time to think and try to solve those unsolved problem after one ACM contest?
No? Oh, you must do this when you want to become a "Big Cattle".
Now you will find that this problem is so familiar:
The greatest common divisor GCD (a, b) of two positive integers a and b, sometimes written (a, b), is the largest divisor common to a and b. For example, (1, 2) =1, (12, 18) =6. (a, b) can be easily found by the Euclidean algorithm. Now I am considering a little more difficult problem:
Given an integer N, please count the number of the integers M (0<M<N) which satisfies (N,M)>1.
This is a simple version of problem “GCD” which you have done in a contest recently,so I name this problem “GCD Again”.If you cannot solve it still,please take a good think about your method of study.
Good Luck!

Input contains multiple test cases. Each test case contains an integers N (1<N<100000000). A test case containing 0 terminates the input and this test case is not to be processed.

For each integers N you should output the number of integers M in one line, and with one line of output for each line in input.

#include <stdio.h>
#include <string.h>
using namespace std;
typedef long long LL;
LL phi(LL x)
{
    LL ans=x;
    for(LL i=2; i*i<=x; i++)
        if(x%i==0)
        {
            ans=ans/i*(i-1);
            while(x%i==0) x/=i;
        }
    if(x>1)
        ans=ans/x*(x-1);
    return ans;
}

int main(){
    LL n;
    while(scanf("%lld",&n)!=EOF,n){
        printf("%lld\n",n-phi(n)-1);
    }
}