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;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using static System.Console;
using static System.Math;
 
public class atc{
    static public int mod = 1000000007;
    static public string al = "abcdefghijklmnopqrstuvwxyz";
    public static void Main(){
        // 方針
        //
        var n = rint();
        var list = new List<List<int>>();
        var root = new int[n];
        for(int i=0;i<n;i++){
            list.Add(new List<int>());
            root[i] = -1;
        }
        
        for(int i=0;i<n;i++){
            var a = inta();
            if(a[1] == 0) continue;
            for(int j=2;j<a[1]+2;j++){
                list[a[0]].Add(a[j]);
                root[a[j]] = a[0];
            }
        }
        
        var ne = -1;
        var depth = new int[n];
        for(int i=0;i<n;i++){
            if(root[i] == -1){
                ne = i;
                break;
            }
        }
        var queue = new Queue<int>();
        queue.Enqueue(ne);
        while(queue.Count > 0){
            var de = queue.Dequeue();
            for(int i=0;i<list[de].Count;i++){
                depth[list[de][i]] = depth[de]+1;
                queue.Enqueue(list[de][i]);
            }
        }
        
        for(int i=0;i<n;i++){
            Write($"node {i}: parent = {root[i]}, depth = {depth[i]}, ");
            if(root[i] == -1){
                Write("root, [");
            }else if(list[i].Count > 0){
                Write("internal node, [");
            }else{
                Write("leaf, [");
            }
            Write(string.Join(", ",list[i]));
            WriteLine("]");
        }
	
	
    }
    public static void swap(ref int a,ref int b){int temp = a;a= b;b = temp;}
    static void charswap(ref char a,ref char b){char temp = a;a= b;b = temp;}
    static int ncr(int n,int r){if(n<r)return 0;r = Min(r,n-r);long nn = 1;for(int i=n-r+1;i<=n;i++){nn = nn*i%mod;}long rr = 1;for(int i=1;i<=r;i++){rr = rr*i%mod;}rr = square((int)rr,mod-2);nn = nn * rr %mod;return (int)nn;}
    // a^b mod
    static int square(int a,int b){string binary = Convert.ToString(b,2);int bileng = binary.Length;long a_power = a;long value = 1;for(int i=bileng-1;i>=0;i--){if(binary[i] == '1'){value = value*a_power%mod;}a_power = a_power*a_power%mod;}return (int)value;}
    static int square2(int a,int b){long output = 1;var list = new List<long>();int sh = 1;long n = a;list.Add(a);while(sh < b){sh *= 2;n = n*n%mod;list.Add(n);}for(int i=list.Count-1;i>=0;i--){if(b > sh){b -= sh;sh /= 2;output = output*list[i]%mod;}}return (int)output;}
    //各種読取
    static string rstr(){ return ReadLine(); }
    static int rint(){ return int.Parse(ReadLine()); }
    static long rlong(){ return long.Parse(ReadLine()); }
    static string[] stra(){ return ReadLine().Split(' '); }
    static char[] chara(){ string[] a=stra();string b="";for(int i=0;i<a.Length;i++){b+=a[i];}return b.ToCharArray();}
    static int[] inta(){ string[] read_str_array = ReadLine().Split(' '); int[] return_int_array = new int[read_str_array.Length]; for(int countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_int_array[countup_i] = int.Parse(read_str_array[countup_i]); } return return_int_array; }
    static int[,] inta2(int num_array,int in_array){ int[,] int_array2 = new int[num_array,in_array]; for(int i=0;i<num_array;i++){ int[] temp_array = inta(); for(int j=0;j<in_array;j++){ int_array2[i,j] = temp_array[j]; } } return int_array2; }
    static long[] longa(){ string[] read_str_array = ReadLine().Split(' '); long[] return_long_array = new long[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_long_array[countup_i] = long.Parse(read_str_array[countup_i]); } return return_long_array; }
    static double[] doublea(){ string[] read_str_array = ReadLine().Split(' '); double[] return_double_array = new double[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_double_array[countup_i] = double.Parse(read_str_array[countup_i]); } return return_double_array; }
    // -----------------------------
    static long GCD(long a,long b){ if(a < b){ long temp = a; a = b; b = temp; } if(a % b == 0){ return b; } else{ long temp = b; b = a%b; a = temp; return GCD(a,b); } }
    static long LCM(long a,long b){ return a * b / GCD(a,b); }
    static void WriteArray(int[,] a,int b,int c){for(int i=0;i<b;i++){for(int j=0;j<c;j++){if(j!=0) Write(" ");Write(a[i,j]);}WriteLine();}}
}