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: Accepted)

def calcs(x,y,ax,ay,bx,by) ((ax-x)*(by-y)-(bx-x)*(ay-y))/2.0 end
def inpf() a=gets.chomp.split(" ").map(&:to_f)end
def inps() a=gets.chomp.split(" ")end  
def copy(a) Marshal.load(Marshal.dump(a)) end
def kaizyo(n)(n < 2)? 1 : (2..n).inject(:*) end
def scount(a) b=na(a.max+1);a.each{|n|b[n]+=1};return b end
def na(n,d=0) Array.new(n,d)end
def na2(n,m,d=0) Array.new(n){Array.new(m,d)}end
def na3(n,m,l,d=0) Array.new(n){Array.new(m){Array.new(l,d)}}end
def bit(n) n.to_s(2).split("").map(&:to_i) end
def inp() a=gets.chomp.split(" ").map(&:to_i)end 
def mtree(x,d,p)
  ret = []
  @md = d if(d>@md)
  if(@data[x][1] > 0)
    @data[x][2..(@data[x][1]+1)].each do|nx|
      ret.push(nx)
      (mtree(nx,d+1,x))
    end
    @tab.push([x,d,p,ret])
  else
    @tab.push([x,d,p,[]])
  end
end
@md = 0
@tab = []
n = inp[0]
@data = na(n)
r = []
n.times do|d|
  @data[d] = inp
end
@data.sort!
n.times do|d|
  r.concat(@data[d][2..(@data[d][1]+1)]) if(@data[d][1] > 0)
end
if(r != [] and r != nil)
  x = scount(r).index(0)
else
  x = 0
end 
mtree(x,0,-1)
@tab.sort!
n.times do |d|
  j = "leaf"
  j = "internal node" if(@tab[d][3].size!=0)
  j = "root" if(@tab[d][1] == 0)
  puts "node #{@tab[d][0]}: parent = #{@tab[d][2]}, depth = #{@tab[d][1]}, #{j}, #{@tab[d][3]}"
end
=begin

=end