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):
    result = 1
    x = x % p
    while y > 0:
        if y % 2 == 1:
            result = (result * x) % p
        y = y // 2
        x = (x * x) % p
    return result

def phi(n):
    result = n
    p = 2
    while p * p <= n:
        if n % p == 0:
            while n % p == 0:
                n //= p
            result -= result // p
        p += 1
    if n > 1:
        result -= result // n
    return result

def count_primitive_roots(p):
    phi_p = phi(p)
    primitive_roots = 0
    for x in range(2, p):
        if power(x, phi_p, p) == 1:
            primitive_roots += 1
    return primitive_roots

p = int(input())
result = count_primitive_roots(p)
print(result)
Replace the power and phi functions as provided in my previous response. This code should correctly solve the problem and give you the number of primitive roots modulo p.




User
wrong answer when input = 5
ChatGPT
I apologize for the confusion earlier. The provided code does not handle the condition where x^k % p is not equal to 1 for all k from 1 to phi(p)-1. To check for this condition, you can use a set to keep track of the residue classes generated by each x^k % p, and then check if the set contains all residue classes that are coprime to p.

Here's the corrected Python code:

python
Copy code
def power(x, y, p):
    result = 1
    x = x % p
    while y > 0:
        if y % 2 == 1:
            result = (result * x) % p
        y = y // 2
        x = (x * x) % p
    return result

def phi(n):
    result = n
    p = 2
    while p * p <= n:
        if n % p == 0:
            while n % p == 0:
                n //= p
            result -= result // p
        p += 1
    if n > 1:
        result -= result // n
    return result

def count_primitive_roots(p):
    phi_p = phi(p)
    primitive_roots = 0
    for x in range(2, p):
        is_primitive_root = True
        residues = set()
        for k in range(1, phi_p):
            residue = power(x, k, p)
            residues.add(residue)
            if residue == 1:
                is_primitive_root = False
                break
        if is_primitive_root and len(residues) == phi_p - 1:
            primitive_roots += 1
    return primitive_roots

p = int(input())
result = count_primitive_roots(p)
print(result)