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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {

	static class Node {
		int id;
		int parentId;
		int depth;
		int[] children;
		public Node(int id, int parentId) {
			this.id = id;
			this.parentId = parentId;
			this.children = null;
			this.depth = 0;
		}
		public void setChildlen(int[] childlen) {
			this.children = childlen;
		}
		public void setParentId(int parentId) {
			this.parentId = parentId;
		}
		public void setDepth(int depth) {
			this.depth = depth;
		}
		private String getType() {
			if (depth == 0) {
				return "root";
			}
			if (children == null || children.length == 0) {
				return "leaf";
			}
			return "internal node";
		}		
		public String toString() {
			StringBuilder sb = new StringBuilder();
			sb.append("node " + id + ": ");
			sb.append("parent = " + parentId + ", ");
			sb.append("depth = " + depth + ", ");
			sb.append(getType() + ", ");
			sb.append(Arrays.toString(children));
			return sb.toString();
		}
	}
		
	public static void main(String[] args) throws IOException {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(reader.readLine());
		Node[] nodes = new Node[n];

		for (int i = 0; i < n; i++) {
			
			StringTokenizer st = new StringTokenizer(reader.readLine());
			
			int id = Integer.parseInt(st.nextToken());
			if (nodes[id] == null) {
				nodes[id] = new Node(id, -1);
			} else {
				if (nodes[id].parentId != -1) {
					nodes[id].setDepth(nodes[nodes[id].parentId].depth + 1);
				}
			}
			
			int m = Integer.parseInt(st.nextToken());
			int[] children = new int[m];
			for (int j = 0; j < m; j++) {
				int childId = Integer.parseInt(st.nextToken());
				children[j] = childId;
				if (nodes[childId] == null) {
					nodes[childId] = new Node(childId, id);
				} else {
					nodes[childId].setParentId(id);
				}
				nodes[childId].setDepth(nodes[id].depth + 1);
			}
			nodes[id].setChildlen(children);
		}
		
		for (Node node : nodes) {
			System.out.println(node);
		}
	}
}