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: Accepted)

#!/usr/bin/env python
# -*- coding: utf-8 -*-


import sys


class Node(object):
    __slots__ = ('parent', 'depth', 'type', 'children')

    def __init__(self):
        self.parent = -1
        self.type = ''
        self.depth = 0
        self.children = []


def dfs(node_idx, node_list, d=0):
    node_list[node_idx].depth = d
    for child in node_list[node_idx].children:
        dfs(child, node_list, d=d + 1)
    return None


def print_result(_node):
    print('parent = ' + str(_node.parent) + ',', end=' ')
    print('depth = ' + str(_node.depth) + ',', end=' ')
    print(_node.type + ',', end=' ')
    print(_node.children)
    return None


def generate_tree(_array, node_list):
    for each in _array:
        idx, k, *children = [x for x in each]
        node_list[idx].children = children
        if k:
            node_list[idx].type = 'internal node'
        else:
            node_list[idx].type = 'leaf'
        for child in children:
            node_list[child].parent = idx
    root_idx = [i for i, x in enumerate(node_list) if x.parent == -1][0]
    node_list[root_idx].type = 'root'
    dfs(node_idx=root_idx, node_list=node_list)
    return node_list


if __name__ == '__main__':
    _input = sys.stdin.readlines()
    array_length = int(_input[0])
    array = list(map(lambda x: list(map(int, x.split())), _input[1:]))
    assert len(array) == array_length

    init_node_list = [Node() for _ in range(array_length)]
    ans = generate_tree(_array=array, node_list=init_node_list)
    for key, node in enumerate(ans):
        print('node ' + str(key) + ':', end=' ')
        print_result(_node=node)