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: Wrong Answer)

import java.util.*;

class Node{
	int parent;
	int depth;
	public ArrayList<Integer> children;
	
	Node(){
		children = new ArrayList<Integer>();
		parent = -1;
	}
	
	public int getParent() {
		return parent;
	}
	
	public void setParent(int parent) {
		this.parent = parent;
	}
	
	public int getDepth() {
		return depth;
	}
	
	public void setDepth(int depth) {
		this.depth = depth;
	}
}

public class Main {
	Scanner sc = new Scanner(System.in);
	StringBuilder output = new StringBuilder();
	
	//節点の個数
	int n;
	
	//木構造を示す配列。クラスNodeは自分で定義した。
	Node nodes[];
	
	//題意より以下は入力のための変数
	/*
	 * id 親
	 * k 親の配下の子ノードの数
	 * child 各子ノードの値
	 */
	int id,k,child;
	
	public void input() {
		n = sc.nextInt();
		nodes = new Node[n];
		//インスタンス化
		for(int i=0; i<n; i++) nodes[i] = new Node(); 
	}
	
	public void solve() {
		for(int i=0; i<n; i++) {
			id = sc.nextInt();
			k = sc.nextInt();
			for(int j=0; j<k; j++) {
				child = sc.nextInt();
				nodes[id].children.add(child);
				nodes[child].setParent(id);
			}
		}
		
		for(int i=0; i<n; i++) {
			//ノードが根だった場合深さは0に設定する
			if(nodes[i].getParent() == -1) compute(i,0);
		}
	}
	
	public void compute(int id, int value) {
		nodes[id].setDepth(value);
		//根以降のノードは深さがプラス１した値になるので再帰で呼び出してセットする
		for(int i=0; i<nodes[id].children.size(); i++) compute(nodes[id].children.get(i), value+1);
	}
	
	public void output() {
		for(int i=0; i<n; i++) {
			output.append("node ").append(Integer.toString(i)).append(": parent = ").append(Integer.toString(nodes[i].getParent())).append(", depth = ").append(Integer.toString(nodes[i].getDepth())).append(", ");
			if(nodes[i].getParent() == -1) output.append("root, ");
			else if(nodes[i].children.size() == 0) output.append("leaf, ");
			else output.append("internal node, ");
			
			output.append("[");
			if(nodes[i].children.size() > 0) {
				output.append(Integer.toString(nodes[i].children.get(0)));
				for(int j=0; j<nodes[i].children.size(); j++) output.append(", ").append(Integer.toString(nodes[i].children.get(j)));
			}
			output.append("]\n");
		}
		System.out.print( output.toString() );
	}
	
	public static void main(String[] args) {
		Main obj = new Main();
		obj.input();
		obj.solve();
		obj.output();
	}
}