# Python Program to evaluate
# Mobius def M(N) = 1 if N = 1
# M(N) = 0 if any prime factor
# of N is contained twice
# M(N) = (-1)^(no of distinct
# prime factors)
import math

# def to check if n
# is prime or not
def isPrime(n) :

    if (n < 2) :
        return False
    for i in range(2, n + 1) :
        if (n % i == 0) :
            return False
        i = i * i
    return True

def mobius(n) :

    p = 0

    # Handling 2 separately
    if (n % 2 == 0) :

        n = int(n / 2)
        p = p + 1

        # If 2^2 also
        # divides N
        if (n % 2 == 0) :
            return 0


    # Check for all
    # other prime factors
    for i in range(3, int(math.sqrt(n)) + 1) :

        # If i divides n
        if (n % i == 0) :

            n = int(n / i)
            p = p + 1

            # If i^2 also
            # divides N
            if (n % i == 0) :
                return 0
        i = i + 2

    if(p % 2 == 0) :
        return -1
    else :
        return 1

# Driver Code
print("Mobius numbers from 1..99:")

for i in range(1, 100):
  print(f"{mobius(i):>4}", end = '')
# This code is contributed by
# Manish Shaw(manishshaw1)