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 JAVA (judge verdict: Accepted)

import java.util.*;
import java.io.*;
public class Main {
	static final int NONUM=-1;
	
	public static void main(String[] args)throws IOException{
		final Scanner sc=new Scanner(System.in);
		int a=sc.nextInt();
		NODE[] n = new NODE[a];
		
		for(int i=0;i<n.length;i++){
			int id=sc.nextInt();
			n[id]=new NODE();
			n[id].id=id;
			n[id].k=sc.nextInt();
			n[id].p=NONUM;
			n[id].t="leaf";
			if(n[id].k>0){
				n[id].c=new int[n[id].k];
				for(int j=0;j<n[id].k;j++){
					n[id].c[j]=sc.nextInt();
				}
				n[id].t="internal node";
			}
		}
		
	     for(int i=0; i<a; i++){
	    	 for(int k=0; k<n[i].k; k++){
	    		 n[n[i].c[k]].p = n[i].id;
	         }
	     }
		
		for(int i=0;i<a;i++){
			if(n[i].p==NONUM){
				n[i].t="root";
				for(int j=0;j<n[i].k;j++){
				getdepth(n,n[i].c[j],1);
				}
			}
		}
		
		for(int i=0;i<a;i++){
			System.out.print("node "+n[i].id+": parent = "+n[i].p+ ", depth = "+n[i].d+", "+n[i].t+", [");
			for(int j=0;j<n[i].k;j++){
				if(j==0) System.out.print(n[i].c[j]);
				else System.out.print(", "+n[i].c[j]);
			}
			System.out.println("]");
		}
	}
	
	static void getdepth(NODE[] n,int i,int d){

		n[i].d=d;
		d++;
		for(int j=0;j<n[i].k;j++){
			getdepth(n,n[i].c[j],d);
		}
	}		
}

class NODE{
	int id;
	int k;
	int[] c;
	int p;
	int d;
	String t; 
}