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.
'''
# Submitted By [email protected]
Don't Copy This Code, Try To Solve It By Yourself
'''
# Problem Name = "Cows and Primitive Roots"
# Class: A

import sys
input = sys.stdin.readline
def print(*args, end='\n', sep=' ') -> None:
    sys.stdout.write(sep.join(map(str, args)) + end)

def is_prime(x):
    for i in range(2, x):
        if x%i==0:
            return False
    return True

def Solve():
    p=int(input())
    c = p//2
    if is_prime(c) and c>2:
        print(c-1)
    elif c%3==0:
        c-=c//3
        print(c)
    else:
        print(c)


if __name__ == "__main__":
    Solve()