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: WA: Presentation Error)

import java.io.*;
import java.util.*;


public class Main{
    static int count =0;
    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String str;
     
        str = br.readLine();
        int n = Integer.parseInt(str);

        Nodes[] line = new Nodes[n];
        
        StringBuilder ans = new StringBuilder();
        for(int i=0;i<n;i++){
            line[i] = new Nodes();
            str = br.readLine();
            String[] tmp = str.split("\\s");

            int[] row = new int[tmp.length];//rowは節点の数字
            for(int j=0;j<tmp.length;j++){
                row[j] = Integer.parseInt(tmp[j]);
            }
            
            //Fisrt:determin ID
            line[i].ID=row[0];

            //Second:detrmin Ck
            if(row[1] != 0){
                line[i].Ck = new int[row[1]];
                for(int j=0;j<row[1];j++){
                    line[i].Ck[j] = row[2+j];
                }
            }
        }

        //Third :determin P ここがオカシイ！！
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(i != j && line[j].Ck != null ){
                    for(int k=0;k<line[j].Ck.length;k++){
                        if(line[i].ID == line[j].Ck[k]){
                            line[i].P = line[j].ID;

                        }
                    }
                }
            }
        }
/*
        for(int i=0;i<n;i++){
            System.out.print(line[i].ID);
            System.out.print(" ,");
            System.out.print(line[i].P);
            System.out.print(" ,");
            if(line[i].Ck != null){
                for(int j=0;j<line[i].Ck.length;j++){
                    System.out.print(line[i].Ck[j]);
                    System.out.print(" ,");
                }
            }
            System.out.print("\n");
        }
*/


        //Forth:determin D & Type ,restart from here
        for(int i=0;i<n;i++){
            line[i].D = FindD(line,line[i].P,0);

            if(line[i].P == -1){
                line[i].Type = "root";
            }else if(line[i].Ck == null){
                line[i].Type = "leaf";
            }else{
                line[i].Type = "internal node";
            }
        
            ans.append("node ");
            ans.append(line[i].ID);
            ans.append(": parent = ");
            ans.append(line[i].P);
            ans.append(", depth = ");
            ans.append(line[i].D);
            ans.append(", ");
            ans.append(line[i].Type);
            ans.append(", [");
            if(line[i].Ck != null){
                for(int j=0;j<line[i].Ck.length;j++){
                    ans.append(line[i].Ck[j]);
                    if(j != line[i].Ck.length-1){
                        ans.append(",");
                    }else{
                        ans.append("]");
                        ans.append("\n");
                    }
                }
            }else{
                ans.append("]");
                ans.append("\n");
            }
        }
        
        System.out.print(ans.toString());
    }
    public static int FindD(Nodes[] line,int P,int depth){//depth = 0で入力
        if(P != -1){
            depth++;
            depth = FindD(line,line[P].P,depth);
        }
        return depth;
    }

}

class Nodes{
    int ID;//node
    int P;//親ノード
    int D;//depth
    String Type;//root,internal node,leaf
    int[] Ck;//節点
    Nodes(){
        this.ID = -1;
        this.P =-1;
        this.D =0;
        this.Type = "leaf";
    }
}