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: Runtime Error)

class TreeNode

	attr_accessor :id, :parent, :children

	def initialize(id, parent=nil)
		@id = id
		@parent = parent
		@children = []
	end

	def add_children(*id_list)
		id_list.each { |id| @children << TreeNode.new(id, self) }
		self
	end

	def merge(other)
		if @id == other.id
			@children = other.children.map { |c| c.parent = self; c }
			return self
		end

		result = false
		@children.each do |c|
			result = !!c.merge(other)
			break if result
		end
		result ? self : nil
	end

	def get_node(id)
		return self if id == @id
		result = nil
		@children.each do |c|
			result = c.get_node(id)
			break if result
		end
		result
	end

	def depth
		return 0 unless @parent
		@parent.depth + 1
	end

	def type
		return "root" unless @parent
		return "internal node" unless @children.empty?
		"leaf"
	end

	def children_id_list
		@children.map { |c| c.id }
	end

	def to_s
		s  = "node #{@id}: "
		s += "parent = #{@parent ? @parent.id : -1}, "
		s += "depth = #{depth}, "
		s += "#{type}, "
		s += "[#{children_id_list.join(", ")}]"
	end

end

n = STDIN.gets.to_i
input_id_list = []
trees = []

n.times do
	id, degree, *children = STDIN.gets.split.map(&:to_i)
	input_id_list << id

	merged = false
	tree = TreeNode.new(id).add_children(*children[0...degree])
	trees.each do |t|
		merged = !!t.merge(tree)
		break if merged
	end

	trees << tree unless merged
end

# マージできる木は必ず後ろにある(でなければ出現時にマージされるため)
# また、最終的には必ず一つになる制約なのでinjectで処理
tree = trees.reverse!.inject { |r, t| r.merge(t) }

input_id_list.sort!.each { |id| puts tree.get_node(id) }