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��zS),aTReturn a natural representation of a timedelta or number of seconds.

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        months (bool): If `True`, then a number of months (based on 30.5 days) will be
            used for fuzziness between years.
        minimum_unit (str): The lowest unit that can be used.

    Returns:
        str (str or `value`): A natural representation of the amount of time
            elapsed unless `value` is not datetime.timedelta or cannot be
            converted to int. In that case, a `value` is returned unchanged.

    Raises:
        OverflowError: If `value` is too large to convert to datetime.timedelta.

    Examples
        Compare two timestamps in a custom local timezone::

        import datetime as dt
        from dateutil.tz import gettz

        berlin = gettz("Europe/Berlin")
        now = dt.datetime.now(tz=berlin)
        later = now + dt.timedelta(minutes=30)

        assert naturaldelta(later - now) == "30 minutes"
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        months (bool): If `True`, then a number of months (based on 30.5 days) will be
            used for fuzziness between years.
        minimum_unit (str): The lowest unit that can be used.
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    will be zero. The rational is that if `unit` is the unit of the quotient, we cannot
    represent the remainder because it would require a unit smaller than the
    `minimum_unit`.

    >>> from humanize.time import _quotient_and_remainder, Unit
    >>> _quotient_and_remainder(36, 24, Unit.DAYS, Unit.DAYS, [])
    (1.5, 0)

    If unit is in `suppress`, the quotient will be zero and the remainder will be the
    initial value. The idea is that if we cannot use `unit`, we are forced to use a
    lower unit so we cannot do the division.

    >>> _quotient_and_remainder(36, 24, Unit.DAYS, Unit.HOURS, [Unit.DAYS])
    (0, 36)

    In other case return quotient and remainder as `divmod` would do it.

    >>> _quotient_and_remainder(36, 24, Unit.DAYS, Unit.HOURS, [])
    (1, 12)

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    >>> from humanize.time import _carry, Unit
    >>> _carry(2, 6, 24, Unit.DAYS, Unit.SECONDS, [Unit.DAYS])
    (0, 54)

    If the unit is the minimum unit, `value2` is divided by `ratio` and added to
    `value1` (carry to left). We assume that `value2` has a lower unit so we need to
    carry it to `value1`.

    >>> _carry(2, 6, 24, Unit.DAYS, Unit.DAYS, [])
    (2.25, 0)

    Otherwise, just return the same input:

    >>> _carry(2, 6, 24, Unit.DAYS, Unit.SECONDS, [])
    (2, 6)
    rr/)r�r�r�r�rnr�s      r!�_carryr�ksL��<�x�������&��)�)��x����&�6�E�>�)�)�)��6�>�r#c�`�||vr)tD]}||kr||vr|cS�d}t|���|S)a�Return a minimum unit suitable that is not suppressed.

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    >>> _suitable_minimum_unit(Unit.HOURS, []).name
    'HOURS'

    But if suppressed, find a unit greater than the original one that is not
    suppressed:

    >>> _suitable_minimum_unit(Unit.HOURS, [Unit.HOURS]).name
    'DAYS'

    >>> _suitable_minimum_unit(Unit.HOURS, [Unit.HOURS, Unit.DAYS]).name
    'MONTHS'
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    >>> from humanize.time import _suppress_lower_units, Unit
    >>> [x.name for x in sorted(_suppress_lower_units(Unit.SECONDS, [Unit.DAYS]))]
    ['MICROSECONDS', 'MILLISECONDS', 'DAYS']
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Sd�|dd ���}"|d }#t/d!��|"|#fzS)"a%Return a precise representation of a timedelta.

    ```pycon
    >>> import datetime as dt
    >>> from humanize.time import precisedelta

    >>> delta = dt.timedelta(seconds=3633, days=2, microseconds=123000)
    >>> precisedelta(delta)
    '2 days, 1 hour and 33.12 seconds'

    ```

    A custom `format` can be specified to control how the fractional part
    is represented:

    ```pycon
    >>> precisedelta(delta, format="%0.4f")
    '2 days, 1 hour and 33.1230 seconds'

    ```

    Instead, the `minimum_unit` can be changed to have a better resolution;
    the function will still readjust the unit to use the greatest of the
    units that does not lose precision.

    For example setting microseconds but still representing the date with milliseconds:

    ```pycon
    >>> precisedelta(delta, minimum_unit="microseconds")
    '2 days, 1 hour, 33 seconds and 123 milliseconds'

    ```

    If desired, some units can be suppressed: you will not see them represented and the
    time of the other units will be adjusted to keep representing the same timedelta:

    ```pycon
    >>> precisedelta(delta, suppress=['days'])
    '49 hours and 33.12 seconds'

    ```

    Note that microseconds precision is lost if the seconds and all
    the units below are suppressed:

    ```pycon
    >>> delta = dt.timedelta(seconds=90, microseconds=100)
    >>> precisedelta(delta, suppress=['seconds', 'milliseconds', 'microseconds'])
    '1.50 minutes'

    ```

    If the delta is too small to be represented with the minimum unit,
    a value of zero will be returned:

    ```pycon
    >>> delta = dt.timedelta(seconds=1)
    >>> precisedelta(delta, minimum_unit="minutes")
    '0.02 minutes'

    >>> delta = dt.timedelta(seconds=0.1)
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    '0 minutes'

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