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::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();

    let mut t: Vec<Node> = vec![Node { parent: -1, left: -1, right: -1, depth: -1 }; n];

    let mut l: usize = 0;
    for _ in 0..n {
        let v: usize = iterator.next().unwrap().parse().unwrap();
        let d: usize = iterator.next().unwrap().parse().unwrap();
        for j in 0..d {
            let c: isize = iterator.next().unwrap().parse().unwrap();
            if j == 0 {
                t[v].left = c;
            } else {
                t[l].right = c;
            }
            l = c as usize;
            t[c as usize].parent = v as isize;
        }
    }

    let root = t.iter().position(|it| it.parent == -1).unwrap();

    rec(root, 0, &mut t);

    let mut ans = String::new();
    for i in 0..n {
        let node = &t[i];
        ans += &format!("node {}: parent = {}, depth = {}, ", i, node.parent, node.depth);
        if node.parent == -1 {
            ans += &format!("root, ");
        } else if node.left == -1 {
            ans += &format!("leaf, ");
        } else {
            ans += &format!("internal node, ");
        }

        ans += "[";
        let mut count = 0;
        let mut child = node.left;
        while child != -1 {
            if count > 0 {
                ans += ", ";
            }
            ans += &format!("{}", child);
            child = t[child as usize].right;
            count += 1;
        }
        ans += "]\n";
    }

    return ans.trim().to_string();
}

fn rec(id: usize, depth: isize, tree: &mut Vec<Node>) {
    tree[id].depth = depth;
    if tree[id].right != -1 {
        rec(tree[id].right as usize, depth, tree);
    }
    if tree[id].left != -1 {
        rec(tree[id].left as usize, depth + 1, tree);
    }
}

#[derive(Clone)]
struct Node {
    parent: isize,
    left: isize,
    right: isize,
    depth: isize,
}