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(x, y, p) : 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1
        x = (x * x) % p 
    return res-1
n =int(input())
count = 0
for i in range(2,n):
    if power(i,n-1,n)==0:
        pass
    else:
        continue
    for j in range(2,n-1):
        if power(i,j,n)!=0:
            continue
        else:
            break
    else:
        count+=1
print(count)
    