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    >>> sorted(c)                       # list all unique elements
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    >>> ''.join(sorted(c.elements()))   # list elements with repetitions
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    >>> sum(c.values())                 # total of all counts
    15

    >>> c['a']                          # count of letter 'a'
    5
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    7
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    0

    >>> d = Counter('simsalabim')       # make another counter
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    9

    >>> c.clear()                       # empty the counter
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    Note:  If a count is set to zero or reduced to zero, it will remain
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    >>> c = Counter('aaabbc')
    >>> c['b'] -= 2                     # reduce the count of 'b' by two
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        >>> prime_factors = Counter({2: 2, 3: 3, 17: 1})
        >>> product = 1
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        >>> c.update(d)                 # add elements from another counter
        >>> c['h']                      # four 'h' in which, witch, and watch
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        >>> c.subtract('witch')             # subtract elements from another iterable
        >>> c.subtract(Counter('watch'))    # subtract elements from another counter
        >>> c['h']                          # 2 in which, minus 1 in witch, minus 1 in watch
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        i(RhR{tNotImplementedR\(RRmRRyR�tnewcount((s9/usr/lib/python2.7/site-packages/future/backports/misc.pyt__add__(s	cC@s�t|t�stSt�}xA|j�D]3\}}|||}|dkr)|||<q)q)WxC|j�D]5\}}||krm|dkrmd|||<qmqmW|S(s� Subtract count, but keep only results with positive counts.

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        i(RhR{R�R\(RRmRRyR�R�((s9/usr/lib/python2.7/site-packages/future/backports/misc.pyt__sub__;s	cC@s�t|t�stSt�}xU|j�D]G\}}||}||krQ|n|}|dkr)|||<q)q)Wx?|j�D]1\}}||kr�|dkr�|||<q�q�W|S(s�Union is the maximum of value in either of the input counters.

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