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: Runtime Error)

import java.util.Scanner;

class Main
{

    static int[] a = new int[1000000];
    static int[] b = new int[1000000];
    static int[] c = new int[1000000];
    static int[] de = new int[1000000];
    static int n;

    static void Depth(int a, int i)
    {
        de[a] = i;
        if(c[a] != -1) Depth(c[a],i);
        if(b[a] != -1) Depth(b[a], i + 1);
    }

    static void Print(int u)
    {
        int i,k;
        System.out.print("node " + u + ": parent = " + a[u] + ", depth = " + de[u] + ", ");

        if(a[u] == -1) System.out.print("root, ");
        else if(b[u] == -1) System.out.print("leaf, ");
        else System.out.print("internal node, ");

        System.out.print("[");

     for(i = 0, k = b[u]; k != -1; i++, k = c[k])
                {
                    if(i != 0)System.out.print(", ");
                    System.out.print(k);
                }
        System.out.println("]");
    }

    public static void main(String[] args)
    {
        Scanner In = new Scanner(System.in);
        int d,e,f,g,h;

        n = In.nextInt();

        for(int i = 0; i < n; i++)
            {
                a[i] = -1;
                b[i] = -1;
                c[i] = -1;
            }

        g = h = 1000;

    for(int i = 0; i < n; i++)
            {
                d = In.nextInt();
                e = In.nextInt();
                for(int j = 0; j < e; j++)
                    {
                        f = In.nextInt();

                        if(j == 0)  b[d] = f;
                        else c[g] = f;
                        g = f;
                        a[f] = d;
                    }
            }

    for(int i = 0; i < n; i++)
        {
            if(a[i] == -1) h = i;
        }

    Depth(h,0);

    for(int i = 0; i < n; i++)
        Print(i);

    }
}