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.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

class Node{
	int depth = 0;
	int parent = -1;
	String type = "";
	ArrayList<Integer> children = new ArrayList<Integer>();
} 

public class Main {
	
	private static Scanner src;

	public static void main(String[] args) {
		src = new Scanner(System.in);
		int n = src.nextInt();
		Node[] tree = new Node[n];
		for(int i = 0; i < n; i++) tree[i] = new Node(); 
		for(int i = 0; i < n; i++) {
			int id = src.nextInt(), k = src.nextInt(); 
			for(int j = 0; j < k; j++) {
				Integer child = src.nextInt(); 
				tree[id].children.add(child);
				tree[child].parent = id;
			}
		}
		for(int i = 0; i < n; i++) {
			if(tree[i].parent == -1) {
				Queue<Integer> queue = new LinkedList<Integer>();
				for(Integer child : tree[i].children) queue.add(child);
				int size = tree[i].children.size();
				int d = 1;
				while(!queue.isEmpty()) {
					int child = queue.poll();
					tree[child].depth = d;
					for(Integer j : tree[child].children) queue.add(j);
					size--;
					if(size == 0) {size = queue.size();  d++;}
				}
				break;
			}
		}
		for(int i = 0; i < n; i++) {
			StringBuilder answ = new StringBuilder();
			answ.append("node ").append(i).append(": parent = ").append(tree[i].parent).append(", depth = ")
			.append(tree[i].depth).append(", ");
			if(tree[i].parent == -1) { answ.append("root, ["); } 
			else if(tree[i].children.size() > 0) { answ.append("internal node, ["); }
			else answ.append("leaf, [");
			int l = tree[i].children.size();
			if(l >= 1) {
				for(int j = 0; j < l - 1; j++) {
					answ.append(tree[i].children.get(j)).append(", ");
				}
				answ.append(tree[i].children.get(l - 1)).append("]");
			}else answ.append("]");
			System.out.println(answ);
		}
	}
}