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

class Program
{
    static List<Node> nodes = new List<Node>();

    class Node
    {
        public int parentId = -1;
        public int depth = 0;
        public List<int> childrenIds;
    }

    static void Main(string[] args)
    {
        // 行数を取得
        int n = int.Parse(Console.ReadLine().Split()[0]);
        for (int i = 0; i < n; i++)
        {
            nodes.Add(new Node());
        }

        for (int i = 0; i < n; i++)
        {
            List<int> nums = Console.ReadLine().Split().Select(str => int.Parse(str)).ToList();
            nodes[nums[0]].childrenIds = nums.GetRange(2, nums[1]);
            for (int j = 2; j < nums[1] + 2; j++)
            {
                nodes[nums[j]].parentId = i;
            }
        }

        Node root = nodes.First(node => node.parentId == -1);

        dfs(root, 0);

        // 出力
        for (int i = 0; i < n; i++)
        {
            StringBuilder sb = new StringBuilder();
            sb.Append("node ");
            sb.Append(i.ToString());
            sb.Append(": parent = ");
            sb.Append(nodes[i].parentId.ToString());
            sb.Append(", depth = ");
            sb.Append(nodes[i].depth.ToString());
            if (nodes[i].parentId == -1)
            {
                sb.Append(", root, [");
            }
            else if (nodes[i].childrenIds.Count() > 0)
            {
                sb.Append(", leaf [");
            }
            else
            {
                sb.Append(", internal node [");
            }
            for (int j = 0; j < nodes[i].childrenIds.Count(); j++)
            {
                sb.Append(nodes[i].childrenIds[j].ToString());
                if (j < nodes[i].childrenIds.Count() - 1)
                {
                    sb.Append(", ");
                }
            }
            sb.Append("]");

            System.Console.WriteLine(sb.ToString());
        }
    }

    static void dfs(Node node, int depth)
    {
        foreach (var childId in node.childrenIds)
        {
            node.depth = depth;
            dfs(nodes[childId], depth  +1);
        }
    }
}