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: Memory Limit Exceeded)

import java.util.Scanner;

public class Main {
	public static Scanner sc = new Scanner(System.in);
	public static class node{
		String status,child;
		String [] children;
		int parent=-1,depth,i;
		
		node(){
			this.child = sc.nextLine();
		}
		void spt(){
			this.children = this.child.split(" ");
		}
		
	}
	public static void position(node [] branch,int num){
		int i;
		for(i=0;i<num;i++){
			if(branch[i].parent == -1) branch[i].status = "root";
			else if(branch[i].children[0].equals("0")) branch[i].status = "leaf";
			else branch[i].status = "internal node";
		}
	}
	
	public static void pd(node [] branch,int index){
		int i,j;
		for(i=0;i<index;i++){
			for(j=0;j<Integer.parseInt(branch[i].children[0]);j++){
				branch[Integer.parseInt(branch[i].children[j+1])].parent = i; 
			}
		}
	}
	public static void sd(node [] branch,int index){
		int i,j,c;
		for(i=0;i<index;i++){
			j = i;
			c = 0;
			while(true){
				if(branch[j].parent == -1) break;
				c++;
				j = branch[j].parent;
			}
			branch[i].depth = c;
		}
	}
	
	public static void main(String[] args){
		
		Integer i,j,n = sc.nextInt();
		String buf;
		int bufi;
		node [] tree = new node[n];
		buf = sc.nextLine();
		
		for(i=0;i<n;i++){
			bufi = sc.nextInt();
			sc.skip(" ");
			tree[bufi] = new node();
			tree[bufi].spt();
		}
		pd(tree,n);
		
		position(tree,n);
		sd(tree,n);
		
		/*for(i=0;i<n;i++){
			if(tree[i].children[0].equals("0")) tree[i].children[0] = "";
		}*/
		
		for(i=0;i<n;i++){
			System.out.printf("node %d: parent = %d, depth = %d, %s, [",i,tree[i].parent,tree[i].depth,tree[i].status);
			for(j=1;j<=Integer.parseInt(tree[i].children[0]);j++){
				System.out.print(tree[i].children[j]);
				if(j != (Integer.parseInt(tree[i].children[0]))){
					System.out.print(", ");
				}
			}
			System.out.println("]");
		}
	}

}