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 Rust (judge verdict: Accepted)

use std::io;
use std::io::prelude::*;
use std::io::{BufWriter,Write};

// 左子右兄弟表現を用いる
struct Node {
    id: u32,
    parent: Option<usize>,
    child: Option<usize>,   // 1つ目の子供
    sib: Option<usize>, // 下の兄弟
    depth: u32, 
}

impl Node {
    fn new(id: u32) -> Node {
        Node {
            id: id,
            parent: None,
            child: None,
            sib: None,
            depth: 0,
        }
    }

    fn _rec_calc_depth(nodes: &mut Vec<Node>, cur_depth: u32, cur_id: usize){
        nodes[cur_id].depth = cur_depth;
        if let Some(ch) = nodes[cur_id].child {
            Node::_rec_calc_depth(nodes, cur_depth+1, ch);
        }
        
        if let Some(sib) = nodes[cur_id].sib {
            Node::_rec_calc_depth(nodes, cur_depth, sib);
        }
    }

    fn calc_depth(nodes: &mut Vec<Node>) {
        for i in 0..nodes.len() {
            if nodes[i].parent.is_none() {
                Node::_rec_calc_depth(nodes, 0, i);
                break;
            }
        }
    }

    fn print(nodes: &Vec<Node>) {
        let out = io::stdout();
        let mut out = BufWriter::new(out.lock());

        for node in nodes {
            write!(out, "node {}: ", node.id).unwrap();
            write!(out, "parent = {}, ", match node.parent {
                Some(x) => x.to_string(),
                None => String::from("-1"),
            }).unwrap();
            write!(out, "depth = {}, ", node.depth).unwrap();
            if node.parent.is_none() {
                write!(out, "root, ").unwrap();
            } else if node.child.is_none() {
                write!(out, "leaf, ").unwrap();
            } else {
                write!(out, "internal node, ").unwrap();
            }

            write!(out, "[").unwrap();
            if let Some(c) = node.child {
                write!(out, "{}", c).unwrap();
                let mut sib = nodes[c].sib;
                while let Some(s) = sib {
                    write!(out, ", {}", s).unwrap();
                    sib = nodes[s].sib;
                }
            }
            write!(out, "]\n").unwrap();
        }
    }
}

fn main() {
    let sin = io::stdin();
    let mut l_iter = sin.lock().lines();
    let n: usize = l_iter.next().unwrap().unwrap().trim().parse().unwrap();

    let mut nodes = (0..n)
        .map(|v| Node::new(v as u32))
        .collect::<Vec<Node>>();

    for line in l_iter {
        let l = line.unwrap();
        let vals = l.split_whitespace()
            .map(|v| v.parse::<usize>().unwrap())
            .collect::<Vec<usize>>();
        let idx = vals[0];
        if vals[1] > 0 {    // 子供の数を比較
            let childs = &vals[2..];
            nodes[idx].child = Some(childs[0]);   // 1番目の子供

            // i番目以降の子供をi-1番目の兄弟として登録
            for i in 0..vals[1] {
                nodes[childs[i]].parent = Some(idx);
                if i < vals[1] - 1 {
                    nodes[childs[i]].sib = Some(childs[i+1]);
                }
            }
        }
    }
    Node::calc_depth(&mut nodes);
    Node::print(&nodes);
}