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 C (judge verdict: Accepted)

/*
 *	Argorithms and Data Structures
 *	Lectuire 7 
 *	Rooted Trees
 */

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define N 100000

typedef struct{
	int id_parent;
	int depth;
	int degree;
	char type[15];
	int *id_child;
} Node;

Node *node;

int depth(int);

int main(void){

	int i,j;

	int n_node, id;
	scanf("%d",&n_node);
	node = malloc(sizeof(Node) * n_node);

	/* Initialization */
	for(i=0;i<n_node;i++){
		node[i].id_parent = -1;
		strcpy(node[i].type,"root");
	}
	
	/* Input */
	for(i=0; i<n_node; i++){
		scanf("%d", &id);
		scanf("%d", &node[id].degree);
		
		/*
			(node[a])'s ID is a
			node[a] has (node[a].degree) children
		*/
		node[id].id_child = malloc( sizeof(int) * node[id].degree );
		for(j=0; j<node[id].degree; j++) {
			scanf("%d", &node[id].id_child[j]);		// You are my child.
			node[ node[id].id_child[j] ].id_parent = id;		// I am your father.
		}
	}

	for(id=0; id<n_node; id++){
		/* node[a]'s type is */
		if( node[id].id_parent != -1 ){
			if( node[id].degree > 0 )
				strcpy( node[id].type, "internal node");
			else
				strcpy( node[id].type, "leaf");
		}

		/* node[a]'s depth is */
		node[id].depth = depth( id );
	}

	/* Output */
	for(id=0; id<n_node; id++){
		printf("node %d: parent = %d, depth = %d, %s, [",
				id, node[id].id_parent, node[id].depth, node[id].type );

		for(i=0; i<node[id].degree ; i++){
			printf("%d", node[id].id_child[i]);
			if( i == node[id].degree -1 )
				break;
			printf(", ");
		}
		printf("]");
		free(node[id].id_child);


		//if(id == n_node-1) break;

		printf("\n");
	}
	free(node);


	return 0;
}

int depth( int id ){
	if (node[id].id_parent == -1)
		return 0;
	return 1 + depth( node[id].id_parent ); 
}

//free id_child