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++11 (judge verdict: Accepted)

//  ALDS1-ALDS1_7_A  Tree - Rooted Trees
#include <iostream>
#include <cstdio>
#include <string>
using namespace std;

const int MAX = 100005;
const int NIL = -1;

struct Node
{
	int parent = NIL;
	int left = NIL;
	int right = NIL;
};

Node T[MAX];
int n, D[MAX];

int getDepth(int u)
{
	int depth = 0;
	while (T[u].parent != NIL)
	{
		u = T[u].parent;
		depth++;
	}
	return depth;
}

void rec(int u, int p)
{
	D[u] = p;
	if (T[u].right != NIL ) rec(T[u].right, p); //右の兄弟に同じ深さを設定
	if (T[u].left != NIL ) rec(T[u].left, p+1); //もっとも左の子に自分の深さ＋1を設定
}

string getType(int u)
{
	if (T[u].parent == NIL ) return "root";
	else if (T[u].left == NIL) return "leaf";
	else return "internal node";
}

string getClidren(int u)
{
	string str = "[";
	if(T[u].left!=NIL)
	{
		int node = T[u].left;
		str += to_string(node);
		while (T[node].right != NIL)
		{
			node = T[node].right;
			str += ", " + to_string(node);
		}
	}
	str += "]";

	return str;
}

void printNodeStructs()
{
	for (int i = 0; i < n; i++)
	{
		printf("node %d p=%2d ,l=%2d, r=%2d, d=%2d, dr=%d,type=%s,%s\n",
		i,T[i].parent,T[i].left,T[i].right,getDepth(i),D[i],getType(i).c_str(),getClidren(i).c_str());
	}
}

void print()
{
	for (int i = 0; i < n; i++)
	{
		printf("node %d: parent = %d, depth = %d, %s, %s\n",
		i,T[i].parent,D[i],getType(i).c_str(),getClidren(i).c_str());
	}
}

int main()
{
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		int vertex,depth, c, l;
		cin >> vertex >> depth;
		for (int j = 0; j < depth; j++)
		{
			cin >> c;
			if (j == 0)
			{
				T[vertex].left = c;
			}
			else
			{
				T[l].right = c;
			}
			l = c;
			T[c].parent = vertex;
		}
	}

	int r;
	for (int i = 0; i < n; i++)
	{
		if(T[i].parent == NIL ) {
			r = i;
		}
	}
	rec(r,0);
	
	//printNodeStructs();
	print();
	return 0;
}