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: Accepted)

import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
    
    static final int NIL = -1;

    public static void main(String[] args) {
        
        Scanner sc = new Scanner(System.in);
        
        int n = Integer.parseInt(sc.next());
        int[] parent = new int[n];
        int[] left = new int[n];
        int[] right = new int[n];
        Arrays.fill(parent, NIL);
        Arrays.fill(left, NIL);
        Arrays.fill(right, NIL);
        for(int i = 0; i < n; i++){
            int id = Integer.parseInt(sc.next());
            int k = Integer.parseInt(sc.next());
            int u = id;
            for(int j = 0; j < k; j++){
                int c = Integer.parseInt(sc.next());
                if(j == 0){
                    left[id] = c;
                }else{
                    right[u] = c;
                }
                u = c;
                parent[c] = id;
            }
        }
        
        int[] depth = new int[n];
        int u = 0;
        while(parent[u] != NIL){
            u = parent[u];
        }
        ArrayDeque<Integer> que = new ArrayDeque<>();
        depth[u] = 0;
        if(left[u] != NIL){
            que.add(left[u]);
        }
        while(!que.isEmpty()){
            u = que.remove();
            int d = depth[parent[u]]+1;
            while(u != NIL){
                depth[u] = d;
                if(left[u] != NIL){
                    que.add(left[u]);
                }
                u = right[u];
            }
        }
        
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < n; i++){
            sb.append("node ").append(i);
            sb.append(": parent = ").append(parent[i]);
            sb.append(", depth = ").append(depth[i]);
            if(parent[i] == NIL){
                sb.append(", root, [");
            }else if(left[i] == NIL){
                sb.append(", leaf, [");
            }else{
                sb.append(", internal node, [");
            }
            if(left[i] != NIL){
                sb.append(left[i]);
                int c = left[i];
                while(right[c] != NIL){
                    sb.append(", ").append(right[c]);
                    c = right[c];
                }
            }
            sb.append("]").append(System.lineSeparator());
        }
        System.out.print(sb);
    }
}