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 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        Node[] aN = new Node[n];
        Node nil = new Node(-1, null, null);
        
        for (int i = 0; i < n; i++) {
            aN[i] = new Node(i, nil, null);
        }
        
        for (int i = 0; i < n; i++) {
            int val = sc.nextInt();
            int dim = sc.nextInt();
            Node[] a = new Node[dim];
            for (int j = 0; j < dim; j++) {
                int val2 = sc.nextInt();
                aN[val2].parent = aN[val];
                a[j] = aN[val2];
            }
            aN[val].children = a;
        }
        
        for (int i = 0; i < n; i++) {
            String kind = aN[i].parent == null ? "root" : aN[i].children.length == 0 ? "leaf" : "internal node";
            String children = getStr(aN[i].children);
            
            System.out.print(
                    "node " + aN[i].val + ": parent = " + aN[i].parent.val + ", depth = " + aN[i].calcDepth() + 
                    ", " + kind + ", " + children);
            System.out.println();
        }
    }
    
    private static String getStr(Node[] a) {
        // TODO 自動生成されたメソッド・スタブ
        StringBuilder sb = new StringBuilder();
        sb.append("[");
        
        for (int i = 0; i < a.length; i++) {
            if (i != 0) {
                sb.append(", ");
            }
            sb.append(a[i].val);
        }
        sb.append("]");
        return sb.toString();
    }

    public static class Node {
        int val;
        Node parent;
        Node[] children;
        public Node(int val, Node parent, Node[] children) {
            this.val = val;
            this.parent = parent;
            this.children = children;
        }
        
        public int calcDepth() { 
            int depth = 0;
            Node p = this.parent;
            while(p.val != -1) {
                depth++;
                p = p.parent;
            }
            return depth;
        }
    }
}