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 Ruby (judge verdict: Wrong Answer)

class Node
  attr_accessor :id, :parent, :depth, :type, :child

  def initialize(id, parent, depth, type)
    @id = id
    @parent = parent
    @depth  = depth
    @type   = type
    @child  = []
  end

  def add_child(children)
    unless children.nil?
      children.each do |child|
        @child.push(child)
      end
    end
  end

end

ID = 0
K = 1
CHILD = 2

num_of_nodes = STDIN.gets.to_i
nodes_input = STDIN.read.split("\n")
tree = Array.new(num_of_nodes,nil)


nodes_input.each do |line|
  node = line.split(" ").map{ |i| i.to_i }
  #p node
  if tree[node[ID]].nil?
    # ???o??m?[?h ( ?????????? root ???????o?^?j
    tree[node[ID]]=Node.new(node[ID],-1,0,'root')
  end
  if node[K]>0
    # ?q?m?[?h?????
    tree[node[ID]].add_child(node[CHILD..-1])
    if tree[node[ID]].parent != -1
      # ?e????????? internal node ???m??
      tree[node[ID]].type = 'internal node'
    end
    # ?q?m?[?h?????X?V
    node[CHILD..-1].each do |child|
      if tree[child].nil?
        # ???o????q?m?[?h(?????????? leaf ???????o?^)
        tree[child]=Node.new(child,node[ID],tree[node[ID]].depth+1,'leaf')
      else
        tree[child].parent = node[ID]
        tree[child].depth  = tree[node[ID]].depth+1
        if tree[child].child.empty?
          tree[child].type = 'leaf'
        else
          tree[child].type = 'internal node'
        end
      end
    end
  end
end

tree.each do |node|
  print "node #{node.id}: parent = #{node.parent}, depth = #{node.depth}, #{node.type}, [#{node.child.join(' ')}]\n"
end