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.Scanner;

public class Main {

	private static Node[] nodes;
	private static int[] D;

	public static void main(String[] args) {
		// TODO ?????????????????????????????????????????????
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();

		nodes = new Node[n];
		StringBuilder[] children = new StringBuilder[n];

		for (int i = 0; i < n; i++) {
			nodes[i] = new Node(-1, -1, -1);
		}

		for (int i = 0; i < n; i++) {
			int node = sc.nextInt();
			int Cnum = sc.nextInt();

			int[] Ckari = new int[Cnum];

			children[node] = new StringBuilder("[");

			for (int j = 0; j < Cnum; j++) {
				Ckari[j] = sc.nextInt();

				if (j == 0) {
					nodes[node].left = Ckari[j];
				} else {
					nodes[Ckari[j - 1]].right = Ckari[j];
				}

				nodes[Ckari[j]].parent = node;

				children[node].append(Ckari[j]);
				if (j != Cnum - 1) {
					children[node].append(", ");
				}
			}

			children[node].append("]");
		}

//		int rootJudge = -1;
		for (int i = 0; i < n; i++) {
			if (nodes[i].parent == -1) {
//				rootJudge = i;
				break;
			}
		}

		D = new int[n];
		
		for (int i = 0; i < n; i++) {
			setDepth(i);
		}


		for (int i1 = 0; i1 < n; i1++) {
			StringBuilder kari = new StringBuilder();
			kari.append("node ");
			kari.append(i1);
			kari.append(": parent = ");
			kari.append(nodes[i1].parent);
			kari.append(", depth = ");
			kari.append(D[i1]);
			kari.append(", ");
			if (nodes[i1].parent == -1) {
				kari.append("root, ");
			} else if (nodes[i1].left == -1) {
				kari.append("leaf, ");
			} else {
				kari.append("internal node, ");
			}
			kari.append(children[i1]);
			System.out.println(kari);

		}

	}

	// public static void setDepth(int u, int p) {
	// D[u] = p;
	//
	// if (nodes[u].right != -1) {
	// setDepth(nodes[u].right, p);
	// }
	// if (nodes[u].left != -1) {
	// setDepth(nodes[u].left, p + 1);
	// }
	//
	// return;
	// }

	public static int setDepth(int u) {
		if (nodes[u].parent == -1) {
			return 0;
		} else {
			return setDepth(nodes[u].parent) + 1;
		}
	}

}

class Node {
	int parent;
	int left;
	int right;

	public Node(int parent, int left, int right) {
		super();
		this.parent = parent;
		this.left = left;
		this.right = right;
	}

}