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.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n  = Integer.parseInt(br.readLine());
        Node[] nodes = new Node[n];
        makeNodeinfo(br,nodes,n);

        int i = 0;
        while(nodes[i].parent != -1 ){
        	i++;
        }
        recDepth(nodes,i,0);

        disp(nodes,n);
	}

	public static void disp(Node[] nodes,int n){
		StringBuilder sb = new StringBuilder();
		for(int i = 0 ; i < n ; i++ ){
			sb.append("node ");
			sb.append(String.valueOf(i));
			sb.append(": ");
			sb.append("parent = ");
			sb.append(String.valueOf(nodes[i].parent));
			sb.append(", ");
			sb.append("depth = ");
			sb.append(String.valueOf(nodes[i].depth));
			sb.append(", ");
			if(nodes[i].parent == -1){
				sb.append("root,");
			}else if(nodes[i].left == -1 ){
				sb.append("leaf,");
			}else{
				sb.append("internal node,");
			}

			if(nodes[i].left != -1){
				sb.append(" [");
				dispChild(nodes,nodes[i].left,sb);
				sb.append("]");
			}else{
				sb.append(" []");
			}

			sb.append("\n");
		}

		sb.delete(sb.length() - 1, sb.length());
		System.out.println(sb.toString());
	}

	public static void dispChild(Node[] nodes,int left,StringBuilder sb){
			int currentChild = left;
			while(currentChild != -1){
				sb.append(String.valueOf(currentChild));
				sb.append(", ");
				currentChild = nodes[currentChild].right;
			}
			sb.delete(sb.length() - 2,sb.length());
	}

	public static void recDepth(Node[] nodes,int i,int depth){
		nodes[i].depth = depth;
		if(nodes[i].left != -1) recDepth(nodes,nodes[i].left,depth + 1);
		if(nodes[i].right != -1) recDepth(nodes,nodes[i].right,depth);
	}

	public static void makeNodeinfo(BufferedReader br,Node[] nodes,int n) throws Exception{
		Node node;
		for(int i = 0 ; i < n ; i++){
			nodes[i] = new Node();
		}
		String[] strs;
		for(int i = 0 ; i < n ; i++ ){
			strs = br.readLine().split(" ");
			int x = Integer.parseInt(strs[1]);
			int current = -1;
			int pre = -1;
			for(int j = 0 ; j < x ; j++){
				current = Integer.parseInt(strs[j + 2]);

				if(j == 0){ //??????????????????
					 nodes[i].left = current;
					 nodes[current].parent = i;
					 pre = current;
				}else{  //????????????
					nodes[pre].right = current;
					nodes[current].parent = i;
					pre = current;
				}
			}
		}
	}

}
 class Node {

	 Node(){
		 this.parent = -1;
		 this.left = -1;
		 this.right = -1;
		 this.depth = -1;
	 }

	 int parent;
	 int left;
	 int right;
	 int depth;
 }