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.
# python 3
"""
"""


def cows_and_primitive_roots(p_int):
    count = 0
    for x in range(1, p_int):
        x_power = [pow(x, i)-1 for i in range(1, p_int)]
        # print(x_power)
        if x_power[-1] % p_int == 0:
            divisible = False
            for each in x_power[:-1]:
                if each % p_int == 0:
                    divisible = True
                    break
            if not divisible:
                count += 1
                # print(x)
    return count


if __name__ == "__main__":
    """
    Inside of this is the test. 
    Outside is the API
    """
    p = int(input())

    print(cows_and_primitive_roots(p))
