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.Linq;
namespace A
{    

    class Depth
    {
        public int dep(int[] parent, int id)
        {
            if (parent[id] ==-1)
            {
                return 0;
            }
            return dep(parent, parent[id]) + 1;
        }
    }

    class A
    {
        static void Main(string[] args)
        {
            string input1;
            input1 = Console.ReadLine();
            int n = int.Parse(input1), i, j;
            Depth dep = new Depth();
            int[] data = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
            int[] node = new int[n];
            int[] k = new int[n];
            int[ , ] number = new int[n+1, 100000];
            int[] parent = new int[n];
            int[] depth = new int[n];
            int id;
            for (i=0; i<n; i++)
            {
                parent[i] = -1;
                node[i] = i;
            }
            string[] type = new string[n];
            for (i = 0; i < n; i++)
            {

                id= data[0];
                k[id] = data[1];
                if (k[id]==0)
                {
                    type[id] = "leaf";
                }
                else
                {
                    for (j=0; j<k[id]; j++)
                    {
                        number[id, j] = data[2+j];
                    }
                }

            }

            //深さ計算・親番号判定
            for (i = 0; i < n; i++)
            {
                for (j = 0; j < k[i]; j++)
                {
                    
                    parent[number[i, j]] = node[i];
                }
            }
            for (i = 0; i < n; i++)
            {
                depth[i] = dep.dep(parent, i);
            }

            //表示
            for (i = 0; i < n; i++)
            {              
                if (depth[i]==0)
                {
                    type[i] = "root";
                }
                else if (type[i]!="leaf")
                {
                    type[i] = "internal node";
                }
                Console.Write("node " + node[i] + ": parent = " + parent[i] + ", depth = " + depth[i] + ", " + type[i] + ", [");
                for (j=0; j<k[i]; j++)
                {
                    if (j!=0)
                    {
                        Console.Write(", ");
                    }
                    Console.Write(number[i, j]);
                }
                Console.WriteLine("]");
            }
        }
    }
}