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 C# (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;

/// <summary>
/// 根付き木
/// </summary>
namespace ALDS1_7_A
{
    class Program
    {
        static void Main(string[] args)
        {
            string line;
            int[] vals;

            //入力
            line = Console.ReadLine().Trim();
            var nodeCount = int.Parse(line);

            var nodes = new Dictionary<int, Node>();
            for (int i = 0; i < nodeCount; i++)
            {
                line = Console.ReadLine().Trim();
                vals = line.Split(" ".ToCharArray(), StringSplitOptions.RemoveEmptyEntries).Select(e => int.Parse(e)).ToArray();

                Node node;
                if (!nodes.TryGetValue(vals[0], out node))
                {
                    node = new Node() { Value = vals[0] };
                    nodes.Add(vals[0], node);
                }
                for (int k = 0; k < vals[1]; k++)
                {
                    Node child;
                    if (!nodes.TryGetValue(vals[k + 2], out child))
                    {
                        child = new Node() { Value = vals[k + 2] };
                        nodes.Add(vals[k + 2], child);
                    }
                    node.AddChild(child);
                }
            }

            // 出力
            foreach (var item in nodes.OrderBy(x => x.Key).Select(x => x.Value))
            {
                Console.WriteLine(item.ToString());
            }
        }
    }


    class Node
    {
        public Node Root { get; set; } = null;
        public int Value { get; set; } = 0;
        public List<Node> Children { get; } = new List<Node>();

        public void AddChild(Node child)
        {
            child.Root = this;
            Children.Add(child);
        }
        public string NodeType()
        {
            if (Root == null)
            {
                return "root";
            }
            if (Children.Any())
            {
                return "internal node";
            }
            return "leaf";
        }
        public int Depth()
        {
            if (Root == null)
            {
                return 0;
            }
            return Root.Depth() + 1;
        }

        public override string ToString()
        {
            return $"node {Value}: parent = {(Root != null ? Root.Value : -1)}, depth = {Depth()}, {NodeType()}, [{String.Join(", ", Children.Select(x => x.Value).ToArray()) }]";
        }
    }
}