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: Time Limit Exceeded)

/*id�̃\�[�g���K�v*/


#include<stdio.h>
#include<string.h>
#define N 100000

typedef struct{

	int id;
	int p;
	int d;
	int k;
	char type[20];
	int c[20];
}node;

node no[N];

int main()
{
	int i,j,n,l=0,m,tmp;
  
	while(1)
		scanf("%d",&n);

	for(i = 0;i < n; i++)
	{
		no[i].p = -1;
		no[i].d = 1;
		scanf("%d%d",&no[i].id,&no[i].k);
		for(j = 0; j < no[i].k; j++) scanf("%d",&no[i].c[j]);
	}

	for(i = 0;i < n;i++)
	{
		for(j = i+1;j < n;j++)
		{
			if(no[i].id > no[j].id)
			{
				tmp = no[i].id;
				no[i].id = no[j].id;
				no[j].id = tmp;
				
				tmp = no[i].k;
				no[i].k = no[j].k;
				no[j].k = tmp;


				if(no[i].k - no[j].k >= 0) l = no[i].k;
				else l = no[j].k;

				for(m = 0;m < l;m++)
				{
					tmp = no[i].c[m];
					no[i].c[m] = no[j].c[m];
					no[j].c[m] = tmp;
				}
			}
		}
	}

	for(i=0;i<n;i++)
	{
		if(no[i].k != 0) strcpy(no[i].type,"internal node");
		else strcpy(no[i].type,"leaf");
	}

	for(i = 0; i < n; i++)
	{
		for(j = 0; j < n; j++)
			for(m = 0; m < no[j].k; m++)
				if(no[i].id == no[j].c[m]) no[i].p = no[j].id;
	}


	for(i = 0;i < n;i++)
	{
		m = no[i].p;
		if(m == -1) 
		{
			strcpy(no[i].type,"root");
			no[i].d = 0;
			continue;
		}		
		while(no[m].p != -1)
		{
			no[i].d += 1;
			m = no[m].p;
		}
	}

	for(i = 0;i < n;i++)
	{
		printf("node %d: parent = %d, depth = %d, %s, [",no[i].id,no[i].p,no[i].d,no[i].type);
		if(no[i].k == 0) printf("]\n");
		else
		{
			for(j = 0; j < no[i].k - 1; j++) printf("%d, ",no[i].c[j]);
			printf("%d]\n",no[i].c[j]);
		}
	}

 	 return 0;
}