Primitive Roots


Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 3462

 

Accepted: 2025


Description


We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (x i mod p) | 1 <= i <= p-1 } is equal to { 1, ..., p-1 }. For example, the consecutive powers of 3 modulo 7 are 3, 2, 6, 4, 5, 1, and thus 3 is a primitive root modulo 7. 
Write a program which given any odd prime 3 <= p < 65536 outputs the number of primitive roots modulo p. 


Input


Each line of the input contains an odd prime numbers p. Input is terminated by the end-of-file seperator.


Output


For each p, print a single number that gives the number of primitive roots in a single line.


Sample Input


23 31 79


Sample Output

10
8
#include<stdio.h>
int euler(int n)
{
  int ans=n;
  for(int i=2;i*i<=n;i++)
  {
    if(n%i==0)
    {
      ans=ans/i*(i-1);
      n/=i;
    }
    while(n%i==0)
    {
      n/=i;
    }
  }
  if(n>1)
      return ans/n*(n-1);
  else
      return ans; 
}
int main()
{
  int n;
  while(scanf("%d",&n)!=EOF)
  {
    printf("%d\n",euler(n-1));//或者euler(euler(n))也对。 
  } 
}