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.
t= int(input())
from math import sqrt
def phi(n) : 
  
    result = n 
    p = 2
    while(p * p<= n) : 
        if (n % p == 0) : 
            while (n % p == 0) : 
                n = n // p 
            result = result * (1.0 - (1.0 / (float) (p))) 
        p = p + 1
    if (n > 1) : 
        result = result * (1.0 - (1.0 / (float)(n))) 
   
    return (int)(result) 
for _ in range(t):
    n = int(input())
    print(phi(n-1))