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.
from math import gcd , lcm , sqrt , floor , ceil
MOD = int(1e9)+7

# fastPower
def fast_power(base , pow , res = 1):
  while pow:
    if pow&1:res = (res*base)%MOD
    base = (base*base)%MOD
    pow//=2
  return res

def solve():
  ans = 0
  for x in range(1,p):
    divisible = False
    for j in range(1,p-1):
      if (fast_power(x,j)-1)%p == 0:
        divisible = True
        break
    if not divisible and (fast_power(x,p-1)-1)%p==0:
      ans+=1
  print(ans)

if __name__ == '__main__':
  p = int(input())
  solve()