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.
p=int(input())

def fac2(n):
    factors = []
    i=2
    while i*i<=n:
        while n%i==0:
            factors.append(i)
            n//=i
        i+=1
    if n>1:
        factors.append(n)
    return factors

x=2
y=1
c=0
z=0
while x<p:
    c+=1
    y=y*x
    if p in fac2(y-1):
        if c==p-1:
            c=0
            z=z+1
            x=x+1
            continue
        else:
            c=0
            x=x+1
            y=1
            continue
print(z)
