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from math import sqrt



def sieve(n):

    """

        *   A fast implementation of the prime sieve algorithm. Finds all prime numbers up to a given upper bound.



        *   Params:

            *   n: The upper bound.

        *   Return:

            *   A list of all prime numbers in [0, n].

        *   Implementation:

            *   Implements the sieve of Erasthotenes with a few optimisations for speed.

            *   We only delete multiples of numbers up to `int(sqrt(n))`. This suffices to find all prime numbers.

            *   We manipulate a list of boolean values, indexed by the number itself with the values denoting its primality. This is much faster than deleting elements in the list.

    """

    if n < 2:   return 

    prime = [True]*(n + 1)      #Initialise the list.

    rt = int(sqrt(n))+1

    for x in range(2, rt):      #We only need to seive multiples of numbers <= to `n`.

        if prime[x]:

            for c in range(x*x, n + 1, x):      prime[c] = False

    return [idx for idx, p in enumerate(prime) if p][2:]

from math import sqrt



def sieve(n):

    """

        *   A fast implementation of the prime sieve algorithm. Finds all prime numbers up to a given upper bound.



        *   Params:

            *   n: The upper bound.

        *   Return:

            *   A list of all prime numbers in [0, n].

        *   Implementation:

            *   Implements the sieve of Erasthotenes with a few optimisations for speed.

            *   We only delete multiples of numbers up to `int(sqrt(n))`. This suffices to find all prime numbers.

            *   We manipulate a list of boolean values, indexed by the number itself with the values denoting its primality. This is much faster than deleting elements in the list.

    """

    if n < 2:   return 

    prime = [True]*(n + 1)      #Initialise the list.

    rt = int(sqrt(n))+1

    for x in range(2, rt):      #We only need to seive multiples of numbers <= to `n`.

        if prime[x]:

            for c in range(x*x, n + 1, x):      prime[c] = False

    return [idx for idx, p in enumerate(prime) if p][2:]

from math import sqrt



def sieve(n):

    """

        *   A fast implementation of the prime sieve algorithm. Finds all prime numbers up to a given upper bound.



        *   Params:

            *   n: The upper bound.

        *   Return:

            *   A list of all prime numbers in [0, n].

        *   Implementation:

            *   Implements the sieve of Erasthotenes with a few optimisations for speed.

            *   We only delete multiples of numbers up to `int(sqrt(n))`. This suffices to find all prime numbers.

            *   We manipulate a list of boolean values, indexed by the number itself with the values denoting its primality. This is much faster than deleting elements in the list.

    """

    if n < 2:   return 

    prime = [True]*(n + 1)      #Initialise the list.

    rt = int(sqrt(n))+1

    for x in range(2, rt):      #We only need to seive multiples of numbers <= to `n`.

        if prime[x]:

            for c in range(x*x, n + 1, x):      prime[c] = False

    return [idx for idx, p in enumerate(prime) if p][2:]

from math import sqrt



def sieve(n):

    """

        *   A fast implementation of the prime sieve algorithm. Finds all prime numbers up to a given upper bound.



        *   Params:

            *   n: The upper bound.

        *   Return:

            *   A list of all prime numbers in [0, n].

        *   Implementation:

            *   Implements the sieve of Erasthotenes with a few optimisations for speed.

            *   We only delete multiples of numbers up to `int(sqrt(n))`. This suffices to find all prime numbers.

            *   We manipulate a list of boolean values, indexed by the number itself with the values denoting its primality. This is much faster than deleting elements in the list.

    """

    if n < 2:   return 

    prime = [True]*(n + 1)      #Initialise the list.

    rt = int(sqrt(n))+1

    for x in range(2, rt):      #We only need to seive multiples of numbers <= to `n`.

        if prime[x]:

            for c in range(x*x, n + 1, x):      prime[c] = False

    return [idx for idx, p in enumerate(prime) if p][2:]

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