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 Go (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

type node struct { // 左子右兄弟表現(left-child, right-sibling representation)
	parent, left, right int
}

func nextInt(sc *bufio.Scanner) int {
	sc.Scan()
	t, _ := strconv.Atoi(sc.Text())
	return t
}

func main() {
	var n int
	fmt.Scan(&n)
	nodes = make([]node, n)
	for i := 0; i < n; i++ {
		nodes[i].parent = -1
		nodes[i].left = -1  // 一番左の子ども
		nodes[i].right = -1 // すぐ隣の兄弟
	}
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)
	for i := 0; i < n; i++ {
		me := nextInt(sc)
		numc := nextInt(sc)
		var left int
		for j := 0; j < numc; j++ { // 子どもをスキャンする
			c := nextInt(sc)
			if j == 0 {
				nodes[me].left = c // 最初の子どもをleftにセットする
			} else {
				nodes[left].right = c // 二番目以降の子どもは、leftのrightにセットする
			}
			left = c
			nodes[c].parent = me
		}
	}
	var root int
	for i := 0; i < n; i++ {
		if nodes[i].parent == -1 {
			root = i
		}
	}
	depth = make([]int, n)
	getDepth(root, 0)
	for i := 0; i < n; i++ {
		printNode(i)
	}
}

var depth []int
var nodes []node

func getDepth(u, p int) {
	// fmt.Printf("getDepth() en: u %d, p %d\n", u, p)
	depth[u] = p
	if nodes[u].right != -1 {
		getDepth(nodes[u].right, p)
	}
	if nodes[u].left != -1 {
		getDepth(nodes[u].left, p+1)
	}
}

func printNode(id int) {
	wtr := bufio.NewWriter(os.Stdout)
	fmt.Fprintf(wtr, "node %d: parent = %d, depth = %d, ", id, nodes[id].parent, depth[id])
	if nodes[id].parent == -1 {
		fmt.Fprintf(wtr, "root, ")
	} else if nodes[id].left == -1 {
		fmt.Fprintf(wtr, "leaf, ")
	} else {
		fmt.Fprintf(wtr, "internal node, ")
	}
	fmt.Fprintf(wtr, "[")
	n := nodes[id].left
	var cnt int
	for n != -1 {
		if cnt == 0 {
			fmt.Fprintf(wtr, "%d", n)
		} else {
			fmt.Fprintf(wtr, ", %d", n)
		}
		cnt++
		n = nodes[n].right
	}
	fmt.Fprintf(wtr, "]\n")
	wtr.Flush()
}