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: Wrong Answer)

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

namespace Project
{
	class Node
	{
		public int num;
		public int dim;
		public int[] c;
		public int parent = -1;
		public string type;
		public int depth;
		public Node(int num, int dim, int[] c)
		{
			this.num = num;
			this.dim = dim;
			this.c = new int[dim];
			for (int i = 0; i < dim; i++)
			{
				this.c[i] = c[i];
			}
		}
	}

	class MainClass
	{
		public static void Main(string[] args)
		{
			int n = int.Parse(Console.ReadLine());
			Node[] nodes = new Node[n];
			for (int i = 0; i < n; i++)
			{
				string[] s = Console.ReadLine().Split(' ');
				int num = int.Parse(s[0]);
				int dim = int.Parse(s[1]);
				int[] c = new int[s.Length - 2];
				for (int j = 2; j < s.Length; j++)
				{
					c[j - 2] = int.Parse(s[j]);
				}
				nodes[i] = new Node(num, dim, c);
			}
			for (int i = 0; i < n; i++)
			{
				foreach (int x in nodes[i].c)
				{
					nodes[x].parent = nodes[i].num;
				}
			}
			int root = 0;
			for (int i = 0; i < n; i++)
			{
				if (nodes[i].parent == -1)
				{
					nodes[i].type = "root";
					root = i;
					break;
				}
			}
			ope(nodes, nodes[root], 0);
			Array.Sort(nodes, (a,b) => a.num - b.num);
			for (int i = 0; i < n; i++)
			{
				Console.Write("node {0}: parent = {1}, depth = {2}, {3}, [",nodes[i].num,nodes[i].parent,nodes[i].depth,nodes[i].type);
				string s = null;
				if (nodes[i].c.Count() != 0)
				{
					for (int k = 0; k < nodes[i].c.Count(); k++)
					{
						s += nodes[i].c[k].ToString();
						if (k != nodes[i].c.Count() - 1)
							s += ", ";
					}
				}
				Console.WriteLine(s + "]");
			}
		}

		public static void ope(Node[] nodes, Node node, int depthcnt)
		{
			node.depth = depthcnt;
			if (node.c.Count() != 0 && node.parent != -1)
				node.type = "internal node";
			else if (node.c.Count() == 0 && node.parent != -1)
				node.type = "leaf";
			for (int i = 0; i < node.c.Length; i++)
			{
				ope(nodes, nodes[node.c[i]], depthcnt + 1);
			}
		}
	}
}