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.
def Power(a,b,c):
    if(b==0):
        return 1
    x=a%c
    u=1
    while(b!=0):
        if(b%2==1):
            u=(u*x)%c
        x=(x*x)%c
        b=b//2
    return u



n=int(input())
ans=0
for i in range(1,n):
    ctr=0
    for j in range(1,n-1):
        if((Power(i,j,n)-1)%n)==0:
            ctr=1
            break
    if ctr!=1 and (Power(i,n-1,n)-1)%n==0:
        ans=ans+1
print(ans)
