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.

2
4
0

0
1

/*
HDOJ 1787 欧拉函数 
这里用的是单独求欧拉函数的模版 
*/

#include<iostream>
#include<stdio.h>
using namespace std;
#define N 100000000

int phi(int x)
{
	int i,ans=x;

	for(i=2;i*i<=x;i++)
	{
		if(x%i==0)
		{
			ans-=ans/i;
			//ans=ans/i*(i-1); 两个公式一样的效果 
			while(x%i==0)//i肯定是素数 
				x/=i;
		}
	}
	if(x>1)
		ans=ans/x*(x-1);
	return ans;
}

int main()
{
	int n;
	while(scanf("%d",&n),n)
	{
		printf("%d\n",n-phi(n)-1);
	}
	return 0;
}