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

public class Main {

	static Main main = new Main();

	public static enum TYPE {
		ROOT, INTR, LEAF
	}

	static String TYPE_ROOT = "root";
	static String TYPE_INTR = "internal node";
	static String TYPE_LEAF = "leaf";

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String line = "";

		while ((line = br.readLine()) != null) {

			int n = Integer.parseInt(line);
			Tree tree = main.new Tree();

			for (int i = 0; i < n; i++) {
				line = br.readLine();
				tree.addNode(line);
			}
			for (int i = 0; i < n; i++) {
				Node node = tree.getNode(i, true);
				System.out.println(node.toString());
			}
		}
	}

	class Tree {
		ArrayList<Node> nodes;

		Tree() {
			nodes = new ArrayList<Node>();
		}

		Node getNode(int id, boolean recursive) {
			Node result = null;
			for (Node n : nodes) {
				if (n.id == id)
					return n;
			}
			if (recursive) {
				for (Node n : nodes) {
					result = n.find(id);
					if (result != null)
						break;
				}
			}
			return result;
		}

		void addNode(String info) {
			StringTokenizer st = new StringTokenizer(info);
			Node node;
			int id = Integer.parseInt(st.nextToken());

			// ???????????????????????則?????????????卒蔵
			node = getNode(id, true);
			if (node == null) {
				// ?????則???????????属?????? root ??即??????
				node = new Node(id, null);
			}

			int k = Integer.parseInt(st.nextToken());
			if (k != 0) {
				node.children = new ArrayList<Node>();
				for (int i = 0; i < k; i++) {
					Node cnode;
					int cid = Integer.parseInt(st.nextToken());
					// ??? root ??則?????????????卒蔵
					cnode = getNode(cid, false);
					if (cnode == null) {
						cnode = new Node(cid, node);
					} else {
						cnode.parent = node;// ???????即????
						nodes.remove(cnode);// ?????????????????贈???????????????????????則?????????
					}
					node.children.add(cnode);// ????????????
				}
			}
			// ???????????????????????則
			if (node.parent == null) {
				nodes.add(node);
			}
		}
	}

	class Node {
		int id;
		Node parent;
		ArrayList<Node> children;

		// Node[] children;

		Node(int id, Node parent) {
			this.id = id;
			this.parent = parent;
		}

		Node find(int id) {
			Node result = null;

			// if (this.id == id) {
			// return this;
			// }

			if (children != null) {
				for (Node n : children) {
					// if (n == null)
					// break;
					if (n.id == id)
						return n;
				}
				for (Node n : children) {
					// if (n == null)
					// break;
					result = n.find(id);
					if (result != null)
						break;
				}
			}
			return result;
		}

		int parentId() {
			if (parent == null)
				return -1;
			return parent.id;
		}

		int depth() {
			int result = 0;
			Node pnode = parent;
			while (pnode != null) {
				result++;
				pnode = pnode.parent;
			}
			return result;
		}

		TYPE type() {
			if (parent == null)
				return TYPE.ROOT;
			else if (children == null)
				return TYPE.LEAF;
			else
				return TYPE.INTR;
		}

		String typeByString() {
			String result = "";
			switch (type()) {
			case ROOT:
				result = TYPE_ROOT;
				break;
			case INTR:
				result = TYPE_INTR;
				break;
			case LEAF:
				result = TYPE_LEAF;
				break;
			}
			return result;
		}

		String childIds() {
			String result = "[";
			if (children != null) {
				for (int i = 0; i < children.size(); i++) {
					if (i != children.size() - 1)
						result += children.get(i).id + ", ";
					else
						result += children.get(i).id;
				}
			}
			result += "]";
			return result;
		}

		public String toString() {
			StringBuilder sb = new StringBuilder();
			sb.append("node " + id + ": ");
			sb.append("parent = " + parentId() + ", ");
			sb.append("depth = " + depth() + ", ");
			sb.append(typeByString() + ", ");
			sb.append(childIds());
			return sb.toString();
		}
	}
}