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.io.*;
import java.util.*;

public class Main{
	public static int count = 0;
	public static void main(String[] args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		int[] node = new int[n];
		int[] k = new int[n];
		int[] parent = new int[n];
		int[] depth = new int[n];
		StringBuilder[] c = new StringBuilder[n];
		int sum = 0;
		int[][] childnode = new int[n][];

		for (int i = 0; i<n ;  i++) {
			String[] line = (br.readLine()).split(" ");
			int index = Integer.parseInt(line[0]);
			node[index] = index;
			k[index] = Integer.parseInt(line[1]);
			childnode[index] = new int[k[index]];
			c[index] = new StringBuilder();
			sum = sum + i;
			if (k[index] == 0) {
				continue;
			}else{
				for (int j = 2; j < (k[index]+2) ; j++) {
					childnode[index][j-2] = Integer.parseInt(line[j]);
					sum -= childnode[index][j-2];
					if (j==2) {
						c[index].append(childnode[index][j-2]);
					}else{
						c[index].append(", ").append(childnode[index][j-2]);
					}
				}
			}

		}

		for (int i = 0 ; i<n ; i++ ) {
			if (sum == node[i]) {

				parent[i] = -1;
				for (int j = 0; j<childnode[i].length ; j++ ) {
					parentdepthset(node,parent,depth,childnode,node[i],childnode[i][j]);
				}
			}
		}

		StringBuilder sb = new StringBuilder();
		for (int i = 0; i<n ; i++ ) {
			sb.append("node ").append(node[i]).append(": ");
			sb.append("parent = ").append(parent[i]).append(", ");
			sb.append("depth = ").append(depth[i]).append(", ");
			sb.append(root(parent[i],c[i]));
			sb.append(", [" + c[i] + "]\n");


		}

		System.out.print(sb);
	}






	public static void parentdepthset(int[] node,int[] parent,int[] depth,int[][] childnode,int parentnode,int child){
		for (int i = 0; i<node.length ; i++ ) {
			if (node[i] == child) {
				parent[i] = parentnode;
				depth[i] = (depth[parentnode]+1);
				if (childnode[i].length > 0) {
					for (int j = 0; j<childnode[i].length ; j++ ) {
						parentdepthset(node,parent,depth,childnode,node[i],childnode[i][j]);
					}
				}
			}

		}


	}
	public static String root(int parent,StringBuilder c){
		if (parent == -1) {
			return "root";
		}else if (c.length() == 0) {
			return "leaf";
		}else{
			return "internal node";
		}
	}


	public static StringBuilder toStringBuilder(int[] a){
		StringBuilder sb = new StringBuilder();
		sb.append(a[0]);
		for (int i = 1; i<a.length ; i++ ) {
			sb.append(" ").append(a[i]);
		}

		return sb;
	}

	public static void countingSort(int[] a,int[] b,int max){
		int[] c = new int[max+1];

		for (int j = 0; j<a.length ; j++ ) {
			c[a[j]]++;
		}

		for (int i = 1; i<=max ; i++ ) {
			c[i] = c[i] + c[i-1];
		}

		for (int j = a.length-1; j>=0 ; j-- ) {
			b[c[a[j]]-1] = a[j];
			c[a[j]]--;
		}


	}
}