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.
x = int(input())
i = 2
ar = []
def fac(n):
    global i,ar
    if (n % i == 0):
        ar.append(i)
        fac(n//i)
    else:
        if i<n:
            i += 1
            fac(n)
fac(x-1)

c = 1
nar = set(ar)
for i in nar:
    a = (len(ar) - ar[::-1].index(i)) - (ar.index(i) + 1) + 1
    if a > 1:
        c = c * (i ** (a-1))
        
    else:
        c = c * (i-1)
        
print(c) 


