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)

#include<iostream>
#include<vector>
using namespace std;
const int MAX = 100005;
const int NIL = -1;
enum T { root, leaf, internalNode };
struct Node
{
	int parent;
	int child;
	int sibling;
	T type;
	vector<int> childList;

}Tree[MAX];
int depth[MAX];
int n;
void print(int u)
{
	cout << "node " << u << ": ";
	cout << "parent = " << Tree[u].parent << ", ";
	cout << "depth = " << depth[u] << ", ";
	if (Tree[u].type == root)
		cout << "root, ";
	else if (Tree[u].type == leaf)
		cout << "leaf, ";
	else
		cout << "internal node, ";
	cout << "[";
	for (int i = 0; i < Tree[u].childList.size(); i++)
	{
		if (i)
			cout << ", ";
		cout << Tree[u].childList[i];
	}
	cout << "]" << endl;
}
void getDepth()
{
	for (int i = 0; i < n; i++)
	{
		int d = 0;
		int u = i;
		while (Tree[u].parent != NIL)
		{
			u = Tree[u].parent;
			d++;
		}
		depth[i] = d;
	}
}
int main()
{
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		Tree[i].child = Tree[i].sibling = Tree[i].parent = NIL;
	}
	for (int i = 0; i < n; i++)
	{
		int p, d;
		cin >> p >> d;
		Tree[p].type = (d != 0) ? internalNode : leaf;
		int c, fc;
		for (int i = 0; i < d; i++)
		{
			cin >> c;
			Tree[p].childList.push_back(c);
			if (i == 0)
			{
				fc = c;
				Tree[p].child = c;
			}
			else
				Tree[fc].sibling = c;
			Tree[c].parent = p;
		}
	}
	int r;
	for (int i = 0; i < n; i++)
		if (Tree[i].parent == NIL)
		{
			r = i;
			Tree[i].type = root;
		}
	getDepth();
	for (int i = 0; i < n; i++)
		print(i);
}