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

class Main{

static final PrintWriter out=new PrintWriter(System.out);
static boolean[] hasParent;

static class Node{
int id;
int p;
int rank;
int k;
int[] c;
String type;
Node(int id){
this.id=id; this.p=-1; this.rank=-1; this.type="internal node";
}
}

public static void main(String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int num=Integer.parseInt(br.readLine());
Node[] ns=new Node[num];
hasParent=new boolean[num];
for(int i=0;i<num;i++){
ns[i]=new Node(i);
}
for(int i=0;i<num;i++){
StringTokenizer st=new StringTokenizer(br.readLine());
int id=Integer.parseInt(st.nextToken());
Node n=ns[id];
int k=Integer.parseInt(st.nextToken());
n.k=k;
if(k==0) n.type="leaf";
else{
n.c=new int[k];
for(int j=0;j<k;j++){
int c=Integer.parseInt(st.nextToken());
n.c[j]=c;
ns[c].p=id;
hasParent[c]=true;
}
}
}
for(int i=0;i<num;i++){
if(ns[i].rank<0) ns[i].rank=setRank(i,ns);
}
for(int i=0;i<num;i++){
Node an=ns[i];
StringBuilder sb=new StringBuilder();
sb.append("node ");
sb.append(i);
sb.append(": parent = ");
sb.append(an.p);
sb.append(", depth = ");
sb.append(an.rank);
sb.append(", root, [");
if(an.k!=0){
for(int j=0;j<an.k;j++) sb.append((j==an.k-1)?an.c[j]:an.c[j]+",");
}
sb.append("]");
out.println(sb);
}
out.flush();
}

private static int setRank(int now,Node[] ns){
if(ns[now].rank>=0) return ns[now].rank;
if(!hasParent[now]){
ns[now].type="root";
return ns[now].rank=0;
}
return ns[now].rank=setRank(ns[now].p,ns)+1;
}
}