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.Arrays;
import java.util.Scanner;

public class Main {
	int[] D;
	Node[] T;
	void run() {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		T = new Node[n];
		for(int i = 0; i < n; i++) {
			T[i] = new Node();
		}
		
		int l = 0;
		for(int i = 0; i < n; i++) {
			T[i].parent = -1;
			T[i].left = -1;
			T[i].right = -1;
		}
		
		for(int i = 0; i < n; i++) {
			int id = sc.nextInt();
			int k = sc.nextInt();
			for(int j = 0; j < k; j++) {
				int c = sc.nextInt();
				if(j == 0) T[id].left = c;
				else T[l].right = c;
				l = c;
				T[c].parent = id;
			}
		}
		
		int r = -1;
		for(int i = 0; i < n; i++) {
			if(T[i].parent == -1) r = i;
		}
		D = new int[n];
		rec(r,0);
		
		for(int i = 0; i < n; i++) {
			StringBuilder sb = new StringBuilder();
			sb.append("node ");
			sb.append(i);
			sb.append(": ");
			sb.append("parent = ");
			sb.append(T[i].parent);
			sb.append(", ");
			sb.append("depth = ");
			sb.append(D[i]);
			sb.append(", ");
			if(T[i].parent == -1) sb.append("root, [");
			else if(T[i].left == -1) sb.append("leaf, [");
			else sb.append("internal node, [");
			String s = sb.toString();
			System.out.println(s);
			
			for(int c = T[i].left, j = 0; c != -1; c =T[c].right) {
				if(j != 0) System.out.print(", ");
				System.out.print(c);;
				j++;
			}
			System.out.println("]");
		}
		
		sc.close();
	}
	
	void rec(int u, int p) {
		D[u] = p;
		if(T[u].right != -1) {
			rec(T[u].right, p);
		}
		if(T[u].left != -1) {
			rec(T[u].left, p+1);
		}
	}

	void debug(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

	public static void main(String[] args) {
		new Main().run();
	}
}
class Node {
	int parent, left, right;
}