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 <algorithm>
#include <iostream>
#include <deque>
#include <sstream>
#include <string>
#include <vector>
using namespace std;
string nodetype[3]={"root", "internal node", "leaf"};

struct tree{
	int parent, child, next;
};

int split(deque <int> &tokens,  string &text, char delim=' '){
	int start = 0, end = 0;
	string tmps;
	char *ends;
	tokens.clear();
	for (int i=0; i<(int)text.size(); i++)
		if (text[i] == delim){
		end = i;
		tmps = text.substr(start, end - start);
		tokens.push_back(strtol(tmps.c_str(),&ends,0));
		start = end + 1;
	}
	tmps = text.substr(start);
	tokens.push_back(strtol(tmps.c_str(),&ends,0));
	return 0;
}

int type(tree &node){
	int e;
	if (node.parent == -1) e = 0;
	else if (node.child == -1) e = 2;
	else e = 1;
	return e;
}

int depth(int e, vector <tree> &node){
	int tmp=node[e].parent, d;

	if (tmp == -1) d = 0;
	else d = depth(tmp, node) + 1;

	return d;
}

string children(int i, vector <tree> &node){
	stringstream ss;

	string str = "[";	
	int e = node[i].child;
	while (e!=-1){
		ss << e;
		if (str=="[") str +=  ss.str();
		else str += ", " + ss.str();
		ss.str("");
		e = node[e].next;
	}
	str += "]";

	return str;
}

int main(void){
	int n, e, tmp;
	string text;
	deque <int> buf;
	stringstream ss;

	cin >> n;
	getline(cin, text);
	vector <tree> node(n);

	for (int i=0; i<n; i++){
		node[i].parent = -1;
		node[i].child = -1;
		node[i].next = -1;
	}

	for (int i=0; i<n; i++){
		getline(cin, text);
		split(buf, text);

		e = buf[0];
		buf.pop_front();

		int nc= buf[0];
		buf.pop_front();

		if (nc != 0){
			for (int j=0; j<nc; j++){
				tmp = buf[j];
				if (j==0) node[e].child = tmp;
				else node[buf[j-1]].next = tmp;
				node[tmp].parent = e;
				}
		}
		buf.clear();
	}

	for (int i = 0; i<n; i++){
		ss.str("");
		ss << "node " << i <<": ";
		ss << "parent = " << node[i].parent << ", ";
		ss << "depth = " << depth(i, node) << ", ";
		e = type(node[i]);
		ss << nodetype[e] << ", ";
		ss << children(i, node);
		cout << ss.str() << endl;
	}

	return 0;
}