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)

#rooted_tree
class Node
  attr_accessor :myname,:parent,:children,:depth

  def initialize(my)
    @myname = my
    @parent = -1
    @children = Array.new
    @child_count = 0
  end

  def add_child(child)
    children << child
    child.parent= self
  end

  def print_child
    cc = self.children.length
    printf(", [")
    0.upto(cc-1){|i|
      printf("%s",self.children[i])
      printf(", ") if i < cc-1
    }
    printf("]")
  end

  def print_parent
    printf("%s",self.parent)
  end

  def to_s
    "#@myname"
  end

  def set_depth(d)
    self.depth=d
    for c in self.children
      c.set_depth(d+1)
    end
  end

  def shubetu
    if self.parent == -1
      "root"
    elsif children.length == 0
      "leaf"
    else
      "internal node"
    end
  end

  def describe
    printf("node %s: parent = %s,",self.myname,self.parent)
    printf(" depth = %d,",self.depth)
    printf(" %s",self.shubetu)
    self.print_child
    printf("\n")
  end

end


n = STDIN.gets.split[0].to_i
tree=Array.new
0.upto(n-1){|i|
  tree[i]=Node.new(i)
}


0.upto(n-1){|i|
  s = STDIN.gets.split
  id= s[0].to_i
  cc= s[1].to_i
  p=2
  cc.times{
    #puts tree[s[id].to_i]
    tree[i].add_child(tree[s[p].to_i])
    p += 1
  }
}

def findroot(tree,n)
  0.upto(n-1){|i|
    if tree[i].parent == -1 
      return i
    end
  }
end 

tree[findroot(tree,n)].set_depth(0)

0.upto(n-1){|i|
  tree[i].describe
}