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 JAVA (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
  public static void main(String args[]){
   Scanner sc = new Scanner(System.in);
   int n = sc.nextInt();
   int[][] C = new int[n][10];
   int[] k = new int[n];
   int[] p = new int[n];
   int[] d = new int[n];
   int[] ty = new int[n];/*0:root, 1:internal node, 2:leaf*/
   for(int i = 0;i < n;i++){
      p[i] = -1;
      d[i] = 0;
      ty[i] = 1;
   }
   for(int i = 0;i < n;i++){
     for(int j = 0;j < 10;j++){/*配列の初期化*/
       C[i][j] = 0;
     }
   } 
   for(int i = 0;i < n;i++){/*入力*/
     int id = sc.nextInt();/*節点の番号 */
       k[id] = sc.nextInt();
       
     for(int j = 1;j <= k[id];j++){
      if(k[id] == 0){
        ty[C[id][j]] = 2;        
       }
       C[id][j] = sc.nextInt();
       p[C[id][j]] = id;
       if(p[C[id][j]] == -1){
         d[C[id][j]] = 0;
         ty[C[id][j]] = 0;
       }
       else d[C[id][j]] = d[id] + 1;
     }
   }
   for(int i = 0;i < n;i++){/*出力*/
    /*debug*///System.out.println("k["+i+"] = "+k[i]);
     System.out.print("node "+i+":");
     System.out.print(" parent = "+p[i]+", ");
     System.out.print("depth = "+d[i]+", ");
     if(ty[i] == 0){
      System.out.print("root, ");
     }
     else if(ty[i] == 2){
      System.out.print("leaf, ");
    }
      else if(ty[i] == 1){
        System.out.print("internal node, ");
      }
     System.out.print("[");
     for(int j = 1;j <= k[i];j++){
       System.out.print(C[i][j]);
       if(j != k[i]) System.out.print(", ");
     }
     System.out.print("]");
     System.out.println();
   }
  }
}