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, :child_left, :right_sibling, :depth

  def initialize(id)
    @id = id
    @parent = nil
    @child_left = nil
    @right_sibling = nil
    @depth = 0
  end

  def depth
    if @parent.nil?
      0
    else
      @parent.depth + 1
    end
  end

  def type
    if @parent == nil
      "root"
    elsif @child_left.nil?
      "leaf"
    else
      "internal node"
    end
  end

  def to_s
    ss =
      if @child_left.nil?
        []
      else
        @child_left.sibling_to_s([])
      end
    parent_id =
      if @parent.nil?
        -1
      else
        @parent.id
      end
    "node #{@id}: parent = #{parent_id}, depth = #{depth}, #{type}, [#{ss.join(', ')}]"
  end

  def sibling_to_s(ss)
    ss.push(@id.to_s)
    if @right_sibling != nil
      @right_sibling.sibling_to_s(ss)
    end
    ss
  end
end

def set_depth(tree, u, depth)
  tree[u].depth = depth
  if tree[u].right_sibling != nil
    set_depth(tree, tree[u].right_sibling.id, depth)
  elsif tree[u].child_left != nil
    set_depth(tree, tree[u].child_left.id, depth + 1)
  end
end


n = STDIN.gets.to_i

tree = []
for i in 0..(n - 1)
  as = STDIN.gets.split(' ')
  id = as[0].to_i
  m = as[1].to_i
  children = as.drop(2).map { |x| x.to_i }

  node = tree[id]
  if node.nil?
    node = Node.new(id)
    tree[id] = node
  end

  l = 0
  children.each_with_index { |c, j|
    child_node = Node.new(c)
    tree[c] = child_node
    if j == 0
      tree[id].child_left = child_node
    else
      tree[l].right_sibling = child_node
    end
    l = c
    tree[c].parent = node
  }
end

root = tree.find { |node| node.parent == nil }.id

set_depth(tree, root, 0)

for i in 0..(n-1)
  puts tree[i].to_s
end