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)

// ALDS1_7_A: Rooted Trees

use std::str::FromStr;

fn scan<T: FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().parse().ok().unwrap()
}

fn scan_vec<T: FromStr>() -> Vec<T> {
    let s: String = scan();
    s.split_whitespace().map(|a| T::from_str(a).ok().unwrap()).collect()
}

struct Node { 
    p: Option<usize>, 
    l: Option<usize>, 
    r: Option<usize>, 
}

impl Node {
    fn new() -> Node {
        Node { p: None, l: None, r: None }
    }
}

fn print(d: &Vec<usize>, t: &Vec<Node>, u: usize) {
    print!("node {}: ", u);
    match t[u].p {
        Some(p) => print!("parent = {}, ", p),
        None => print!("parent = -1, "),
    }
    print!("depth = {}, ", d[u]);
    if t[u].p == None {
        print!("root, ");
    } else if t[u].l == None {
        print!("leaf, ");
    } else {
        print!("internal node, ");
    }
    print!("[");
    if let Some(l) = t[u].l {
        let mut c = l;
        print!("{}", c);
        while let Some(r) = t[c].r {
            print!(", {}", r);
            c = r;
        }
    }
    println!("]");
}

fn rec(d: &mut Vec<usize>, t: &Vec<Node>, u: usize, p: usize) {
    d[u] = p;
    if let Some(r) = t[u].r {
        rec(d, t, r, p);
    }
    if let Some(l) = t[u].l {
        rec(d, t, l, p + 1);
    }
}

fn main() {
    let n: usize = scan();
    let mut t: Vec<Node> = Vec::new();

    for _ in 0..n {
        t.push(Node::new())
    }

    for _ in 0..n {
        let a: Vec<usize> = scan_vec();
        let v = a[0];
        let d = a[1];
        let mut l = 0;
        for j in 0..d {
            let c = a[j + 2];
            if j == 0 {
                t[v].l = Some(c);
            } else {
                t[l].r = Some(c);
            }
            l = c;
            t[c].p = Some(v);
        }
    }

    let mut r = 0;
    for i in 0..n {
        if t[i].p == None {
            r = i;
        }
    }

    let mut d: Vec<usize> = vec![0; n];
    rec(&mut d, &t, r, 0);

    for i in 0..n {
        print(&d, &t, i);
    }
}