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::fmt;
type NodeNum = i32;

#[derive(Debug)]
struct Node {
    id: NodeNum,
    parent: Option<NodeNum>,
    depth: usize,
    children: Vec<NodeNum>,
}

impl fmt::Display for Node {
    fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
        let node_type = if self.depth == 0 {
            "root"
        } else if self.children.len() == 0 {
            "leaf"
        } else {
            "internal node"
        };

        let p = match self.parent {
            Some(val) => val,
            None => -1,
        };

        write!(f, "node {}: parent = {:?}, depth = {}, {}, {:?}", self.id, p, self.depth, node_type, self.children)
    }
}

struct Tree {
    nodes: Vec<Node>,
}

impl Tree {
    fn update_node(&mut self, id: NodeNum, childs: Vec<NodeNum>) {
        if childs.len() == 0 { return; }

        let copy_childs = childs.clone();

        self.nodes[id as usize].children = childs;

        for child in copy_childs {
            self.nodes[child as usize].parent = Some(id);
        }
    }

    fn find_root(&self) -> NodeNum {
        let mut current: NodeNum = 0;
        loop {
            match self.nodes[current as usize].parent {
                Some(id) => current = id,
                None => break,
            }
        }
        current
    }

    fn fill_depth(&mut self, root_id: NodeNum, depth: usize) {
        self.nodes[root_id as usize].depth = depth;

        let copy_childs = self.nodes[root_id as usize].children.clone();

        for child in copy_childs {
            self.fill_depth(child, depth + 1);
        }
    }
}

fn main() {
    let n = read::<i32>();

    let mut nodes = Vec::<Node>::new();
    for i in 0..n {
        nodes.push( Node{
            id: i,
            parent: None,
            depth: 0,
            children: Vec::new(),
        });
    }
    let mut tree = Tree { nodes: nodes };


    for _ in 0..n {
        let lines = read_vec::<NodeNum>();

        let childs = if lines.len() < 3 {
            [].to_vec()
        } else {
            lines[2..].to_vec()
        };
        tree.update_node(lines[0], childs);
    }

    let root = tree.find_root();
    tree.fill_depth(root, 0);

    for id_node in 0..n {
        let node = &tree.nodes[id_node as usize];
        println!("{}", node);
    }
}

/// 標準入力から一行を読み取り、指定の型に変換する関数
fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line)
        .ok()
        .expect("Faild to read Vec");
    line.trim().split_whitespace()
        .map(|i| i.parse().ok().unwrap()).collect()
}