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)

import sys
sys.setrecursionlimit(2**20)

class Node:
    def __init__(self, parent, left, right):
        self.parent = parent #親
        self.left = left #一番左の子ノード
        self.right = right #すぐ右の兄弟

def get_depth(T: dict, u: int):
    """
    T: 木構造
    u: ノード番号
    """
    d = 0
    while T[u].parent is not None:
        u = T[u].parent
        d += 1

D = {} #深さを保存

def get_all_depth(T: dict, u: int, p: int):
    """
    一度に全てのノードの深さを再帰的に取得する。
    p: 現在扱う深さ
    """
    D[u] = p
    if T[u].right is not None:
        get_all_depth(T, T[u].right, p) #兄弟から入れていく（深さは同じ）
    if T[u].left is not None:
        get_all_depth(T, T[u].left, p+1) #1つ深く潜る

def ret_children(T: dict, u: int):
    children = []
    c = T[u].left
    while c is not None:
        children.append(c)
        c = T[c].right
    return children

# 答え出力
def print_for_a_node(T: dict, node: int, D: dict):
    node_type = "internal node" if T[node].left is not None else "leaf"
    if T[node].parent is None: #親ノードが無ければ
        parent = -1
        node_type = "root"
    else:
        parent = T[node].parent
    children = ret_children(T, node)
    print(
        f"node {node}: parent = {parent}, depth = {D[node]}, {node_type}, {children}"
    )

# 入力
n = int(input())

T = {k: Node(None, None, None) for k in range(n)}

for _ in range(n):
    tmp = list(map(int, input().split()))
    if tmp[1] == 0: # 子ノードがなければ次へ
        continue
    T[tmp[0]].left = tmp[2] # 一番左の子ノード
    T[tmp[2]].parent = tmp[0] # 子ノードの親（自分）
    prev_sibling = tmp[2]
    for sib in tmp[3:]:
        T[prev_sibling].right = sib
        T[sib].parent = tmp[0]
        prev_sibling = sib

for node in range(n):
    if T[node].parent is None:
        get_all_depth(T, node, 0)
        # Dに深さが格納された
        break
for node in range(n):
    print_for_a_node(T, node, D)