What is the execution sequence of the code in python

Question

Below is the code snippet,

sieve=[0]*(1000001)
def countDivisors():
    global sieve
    print("@2")
    for i in range(1,1000001):
        j=i
        while(j<1000001):
            sieve[j]+=1
            j+=i
class Solution:
    # @param A : list of integers
    # @return a list of integers
    countDivisors()
    def solve(self, A):
        n=len(A)
        ans=[]
        global sieve
        for i in range(n):
            ans.append(sieve[A[i]])
        return ans
                
print("main")          
s=Solution()
print("s declared")
A = [[3, 5, 4, 2],[8, 5, 6, 2]]
for i in A:
    print(s.solve(i))

Output:

@2
main
s declared
[2, 2, 3, 2]
[4, 2, 4, 2]

Why "@2" before "main" is getting printed first. What is the execution sequence of the code and why? How it's different from the below code? In the below code snippet the 'countDivisor()' is called in 'init()'.

sieve=[0]*(1000001)
def countDivisors():
    global sieve
    print("@2")
    for i in range(1,1000001):
        j=i
        while(j<1000001):
            sieve[j]+=1
            j+=i
class Solution:
    # @param A : list of integers
    # @return a list of integers
    def __init__(self):
        countDivisors()
    def solve(self, A):
        n=len(A)
        ans=[]
        global sieve
        for i in range(n):
            ans.append(sieve[A[i]])
        return ans

Answer 1

Unlike a def statement, the body of a class statement is executed in order to prepare the namespace used to define the class. Your class statement is roughly equivalent to

countDivisors()


def solve(self, A):
    n=len(A)
    ans=[]
    global sieve
    for i in range(n):
        ans.append(sieve[A[i]])
    return ans


Solution = type('Solution', (), {'solve': solve})