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 OCaml (judge verdict: Accepted)

module DoublyChainedTree = struct
  let nil = -1

  type t = {
    mutable parent : int;
    mutable child : int;
    mutable bro : int;
    mutable depth : int;
  }

  let make () = { parent = nil; child = nil; bro = nil; depth = nil }
end

module T = DoublyChainedTree

let () =
  let n = read_int () in
  let t = Array.init n (fun _ -> T.make ()) in
  let rec read_tree i =
    if i < n then begin
      match List.map int_of_string (Str.split (Str.regexp " ") (read_line ())) with
      | id :: _ :: tl -> begin
          let rec read_lst j left_child = function
            | [] -> ()
            | c :: cs -> begin
                t.(c).T.parent <- id;
                if j = 0 then t.(id).T.child <- c else t.(left_child).T.bro <- c;
                read_lst (j+1) c cs
              end
          in
          read_lst 0 0 tl;
          read_tree (i + 1)
        end
      | _ -> exit 1
    end
  in
  read_tree 0;
  let rec search_root i =
    if i = n then T.nil
    else if t.(i).T.parent = T.nil then i
    else search_root (i + 1)
  in
  let root = search_root 0 in
  let rec set_depth i p =
    t.(i).T.depth <- p;
    if t.(i).T.bro <> T.nil then set_depth t.(i).T.bro p;
    if t.(i).T.child <> T.nil then set_depth t.(i).T.child (p + 1);
  in
  set_depth root 0;
  let print_node i =
    Printf.printf "node %d: parent = %d, depth = %d, " i t.(i).T.parent t.(i).T.depth;
    print_string (if t.(i).T.parent = T.nil then "root, " else if t.(i).T.child = T.nil then "leaf, " else "internal node, ");
    print_string "[";
    let rec print_children c =
      if c <> T.nil then begin
        if c <> t.(i).T.child then print_string ", ";
        print_int c;
        print_children t.(c).T.bro
      end
    in
    print_children t.(i).T.child;
    print_endline "]"
  in
  let rec print i =
    if i < n then (print_node i; print (i + 1))
  in
  print 0