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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def compute_backoff(attempts, *, factor=5, jitter=True, max_backoff=2000, max_exponent=32):
 """Compute an exponential backoff value based on some number of attempts. Parameters: attempts(int):
 The number of attempts there have been so far. factor(int):
 The number of milliseconds to multiply each backoff by. max_backoff(int):
 The max number of milliseconds to backoff by. max_exponent(int):
 The maximum backoff exponent. Returns: tuple: The new number of attempts and the backoff in milliseconds. """

    exponent = min(attempts, max_exponent)
    backoff = min(factor * 2 ** exponent, max_backoff)
    if jitter:
        backoff /= 2
        backoff = int(backoff + uniform(0, backoff))
    return attempts + 1, backoff