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;

class Node{
  int id;
  int deg;
  int depth;
  Node parent = null;
  Node left = null;
  Node right = null;


  int getParentID(){
    if(parent == null)
      return -1;
    else
      return parent.id;
  }

  String getType(){
    if(parent == null)
      return "root";
    else if(left != null)
      return "internal node";
    else
      return "leaf";
  }

  void printChildren(){
    Node u = this.left;
    System.out.print("[");
    while(u != null){
      if(u.right == null)
        System.out.print(u.id);
      else
        System.out.print(u.id + ", ");
      u = u.right;
    }
    System.out.println("]");
  }

  void printInfo(){
    System.out.print("node " + id + ": ");
    System.out.print("parent = " + this.getParentID() + ", ");
    System.out.print("depth = " + depth + ", ");
    System.out.print(this.getType() + ", ");
    printChildren();
  }

  void setDepth(int d){
    this.depth = d;
    if(this.right != null)
      this.right.setDepth(d);
    if(this.left != null)
      this.left.setDepth(d+1);
  }

  boolean isRoot(){
    if(parent == null)
      return true;
    else
      return false;
  }
}

class Main{
  public static void main(String[] args){
    Scanner stdIn = new Scanner(System.in);
    int n = stdIn.nextInt();
    Node[] node = new Node[n];
    for(int i = 0; i < n; i++)
      node[i] = new Node();

    for(int i = 0; i < n; i++){
      int id = stdIn.nextInt();
      int deg = stdIn.nextInt();
      node[i].id = id;
      node[i].deg = deg;
      int tmp = 0;
      for(int j = 0; j < deg; j++){
        int cid = stdIn.nextInt();
        node[cid].parent = node[id];
        if(j == 0)
          node[id].left = node[cid];
        else
          node[tmp].right = node[cid];
        tmp = cid;
      }
    }

    int root = 0;
    int k = 0;
    while(!node[k].isRoot())k++;
    root = k;
    node[root].setDepth(0);

    for(int i = 0; i < n; i++)
      node[i].printInfo();
  }
}