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.NoSuchElementException;
import java.util.Scanner;
import java.util.StringJoiner;
import java.util.stream.Stream;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        try {
            int n = Integer.parseInt(sc.nextLine());
            Node[] ns = new Node[n];
            for (int i = 0; i < n; i++) {
                ns[i] = new Node();
            }
            while (sc.hasNext()) {
                int[] id_k_cs = Stream.of(sc.nextLine().split(" "))
                        .mapToInt(Integer::parseInt)
                        .toArray();
                int id = id_k_cs[0];
                if (id_k_cs.length > 2) {
                    ns[id].left = id_k_cs[2];
                }
                for (int j = 2; j < id_k_cs.length; j++) {
                    ns[id_k_cs[j]].parent = id_k_cs[0];
                    if (j != id_k_cs.length - 1) {
                        ns[id_k_cs[j]].right = id_k_cs[j + 1];
                    }
                }
            }
            for (int i = 0; i < ns.length; i++) {
                System.out.printf("id, parent, left, right: %d, %d, %d, %d\n",
                        i,
                        ns[i].parent,
                        ns[i].left,
                        ns[i].right);
            }
            for (int i = 0; i < ns.length; i++) {
                int parent = ns[i].parent == null ? -1 : ns[i].parent;
                int depth = getDepth(ns, i);
                String type = getType(ns[i]);
                String cs = getChilds(ns, i);
                System.out.printf(
                        "node %d: parent = %d, depth = %d, %s, %s\n",
                        i,
                        parent,
                        depth,
                        type,
                        cs);
            }
        } catch (NoSuchElementException
                | NumberFormatException e) {
            System.out.println(e);
        }
    }

    private static class Node {
        Node() {}
        Integer parent;
        Integer left;
        Integer right;
    }

    private static int getDepth(Node[] ns, int id) {
        int depth = 0;
        Node node = ns[id];
        while (node.parent != null) {
            depth++;
            node = ns[node.parent];
        }
        return depth;
    }

    private static String getType(Node n) {
        if (n.parent == null) {
            return "root";
        }
        if (n.left == null) {
            return "leaf";
        }
        return "internal node";
    }

    private static String getChilds(Node[] ns, int id) {
        StringJoiner sj = new StringJoiner(", ", "[", "]");
        Integer cid = ns[id].left;
        if (cid == null) {
            return sj.toString();
        }
        Node c = ns[ns[id].left];
        sj.add(cid.toString());
        while (c.right != null) {
            sj.add(c.right.toString());
            c = ns[c.right];
        }
        return sj.toString();
    }
}