Problem:
Rooted Trees

A graph G = (V, E) is a data structure where V is a finite set of vertices and E is a binary relation on V represented by a set of edges. Fig. 1 illustrates an example of a graph (or graphs).

Fig. 1

A free tree is a connnected, acyclic, undirected graph. A rooted tree is a free tree in which one of the vertices is distinguished from the others. A vertex of a rooted tree is called "node."

Your task is to write a program which reports the following information for each node u of a given rooted tree T:

node ID of u

parent of u

depth of u

node type (root, internal node or leaf)

a list of chidlren of u

If the last edge on the path from the root r of a tree T to a node x is (p, x), then p is the parent of x, and x is a child of p. The root is the only node in T with no parent.

A node with no children is an external node or leaf. A nonleaf node is an internal node

The number of children of a node x in a rooted tree T is called the degree of x.

The length of the path from the root r to a node x is the depth of x in T.

Here, the given tree consists of n nodes and evey node has a unique ID from 0 to n-1.

Fig. 2 shows an example of rooted trees where ID of each node is indicated by a number in a circle (node). The example corresponds to the first sample input.

Fig. 2

Input

The first line of the input includes an integer n, the number of nodes of the tree.

In the next n lines, the information of each node u is given in the following format:

id k c1 c2 ... ck

where id is the node ID of u, k is the degree of u, c1 ... ck are node IDs of 1st, ... kth child of u. If the node does not have a child, the k is 0.

Output

Print the information of each node in the following format ordered by IDs:

node id: parent = p, depth = d, type, [c1...ck]

p is ID of its parent. If the node does not have a parent, print -1.

d is depth of the node.

type is a type of nodes represented by a string (root, internal node or leaf). If the root can be considered as a leaf or an internal node, print root.

c1...ck is the list of children as a ordered tree.

Please follow the format presented in a sample output below.

Constraints

1 ≤ n ≤ 100000

Sample Input 1

13
0 3 1 4 10
1 2 2 3
2 0
3 0
4 3 5 6 7
5 0
6 0
7 2 8 9
8 0
9 0
10 2 11 12
11 0
12 0

Sample Output 1

node 0: parent = -1, depth = 0, root, [1, 4, 10]
node 1: parent = 0, depth = 1, internal node, [2, 3]
node 2: parent = 1, depth = 2, leaf, []
node 3: parent = 1, depth = 2, leaf, []
node 4: parent = 0, depth = 1, internal node, [5, 6, 7]
node 5: parent = 4, depth = 2, leaf, []
node 6: parent = 4, depth = 2, leaf, []
node 7: parent = 4, depth = 2, internal node, [8, 9]
node 8: parent = 7, depth = 3, leaf, []
node 9: parent = 7, depth = 3, leaf, []
node 10: parent = 0, depth = 1, internal node, [11, 12]
node 11: parent = 10, depth = 2, leaf, []
node 12: parent = 10, depth = 2, leaf, []

Sample Input 2

4
1 3 3 2 0
0 0
3 0
2 0

Sample Output 2

node 0: parent = 1, depth = 1, leaf, []
node 1: parent = -1, depth = 0, root, [3, 2, 0]
node 2: parent = 1, depth = 1, leaf, []
node 3: parent = 1, depth = 1, leaf, []

Note

You can use a left-child, right-sibling representation to implement a tree which has the following data:

the parent of u

the leftmost child of u

the immediate right sibling of u

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in Python3 (judge verdict: Runtime Error)


class Node(object):
    def __init__(self, parent, left, right):
        self.parent = parent  # ????????????
        self.left = left      # ??????????????????????????????????????????
        self.right = right    # ??????????????????????????????????????´?????????

    def get_depth(self):
        if self.parent == -1:
            return 0
        else:
            depth = 1
            t = Nodes[self.parent]
            while t.parent != -1:
                t = Nodes[t.parent]
                depth += 1
            return depth

    def get_type(self):
        if self.parent == -1:
            return 'root'
        elif self.left == None:
            return 'leaf'
        else:
            return 'internal node'


def decode_node_data(n):
    global Nodes
    num_of_elem = len(n)
    if num_of_elem < 2:
        raise ValueError
    elif num_of_elem == 2:  # leaf
        pass
    else:
        id, num_of_children = int(n[0]), int(n[1])
        children = n[2:]
        Nodes[id].left = int(children[0])
        prev_child = None
        for c in children:
            Nodes[int(c)].parent = id
            if prev_child != None:
                Nodes[prev_child].right = int(c)
            prev_child = int(c)


MAX_NODES = 20
Nodes = [Node(-1, None, None) for _ in range(MAX_NODES)]
if __name__ == '__main__':
    # ??????????????\???
    num_of_nodes = int(input())
    node_data = []
    for _ in range(num_of_nodes):
        node_data.append(input().split(' '))

    # ??¨????§£???
    for n in node_data:
        decode_node_data(n)


    # ???????????¨???
    for node_id in range(num_of_nodes):
        type = Nodes[node_id].get_type()
        children = []
        if type != 'leaf':
            if Nodes[node_id].left != None:
                c = Nodes[node_id].left
                children.append(c)
                while Nodes[c].right != None:
                    c = Nodes[c].right
                    children.append(c)
        print('node {0}: parent = {1}, depth = {2}, {3}, [{4}]'.format(node_id, Nodes[node_id].parent, Nodes[node_id].get_depth(), type,
                                                                       ', '.join(map(str, children))))