human
Write me a the Python program you requested given the following:
Complete the below
def from_nested_tuple(sequence, sensible_relabeling=False):

    ''''Returns the rooted tree corresponding to the given nested tuple.
    The nested tuple representation of a tree is defined
    recursively. The tree with one node and no edges is represented by
    the empty tuple, ``()``. A tree with ``k`` subtrees is represented
    by a tuple of length ``k`` in which each element is the nested tuple
    representation of a subtree.
    Parameters
    sequence : tuple
    A nested tuple representing a rooted tree.
    sensible_relabeling : bool
    Whether to relabel the nodes of the tree so that nodes are
    labeled in increasing order according to their breadth-first
    search order from the root node.
    Returns
    NetworkX graph
    The tree corresponding to the given nested tuple, whose root
    node is node 0. If ``sensible_labeling`` is ``True``, nodes will
    be labeled in breadth-first search order starting from the root
    node.
    Notes
    This function is *not* the inverse of :func:`to_nested_tuple`.
the
    only guarantee is that the rooted trees are isomorphic.
    See also
    to_nested_tuple
    from_prufer_sequence
    Examples
    Sensible relabeling ensures that the nodes are labeled from the root
    starting at 0::
    >>> balanced = (((), ()), ((), ()))
    >>> T = nx.from_nested_tuple(balanced, sensible_relabeling=True)
    >>> edges = [(0, 1), (0, 2), (1, 3), (1, 4), (2, 5), (2, 6)]
    >>> all((u, v) in T.edges() or (v, u) in T.edges() for (u, v) in edges)
    True'
    '''
assistant
#Here is the Python program you requested.
    def _make_tree(sequence):
        'Recursively    creates    a    tree    from    the    given    sequence    of    nested\n                                tuples.\n\n                                This    function    employs    the    :func:`~networkx.tree.join`    function\n                                to    recursively    join    subtrees    into    a    larger    tree.\n\n                                '
        if (len(sequence) == 0):
            return nx.empty_graph(1)
        return nx.tree.join([(_make_tree(child), 0) for child in sequence])
    T = _make_tree(sequence)
    if sensible_relabeling:
        bfs_nodes = chain([0], (v for (u, v) in nx.bfs_edges(T, 0)))
        labels = {v: i for (i, v) in enumerate(bfs_nodes)}
        T = nx.relabel_nodes(T, labels)
    return T