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 C++ (judge verdict: Time Limit Exceeded)

#include <iostream>

using namespace std;

int nil = -1;
int N;


struct Node {
    int id;
    int left, right, parent;
};

Node *nodes;

/*
 * 左子右兄弟表現
 */

int depth(int m) {
    int d = 0;
    while (true) {
        if (nodes[m].parent != nil) {
            d++;
            m = nodes[m].parent;
        } else break;
    }
    return d;
}

string type(int m) {
    if (nodes[m].parent == nil) {
        return  "root";
    } else if (nodes[m].left != nil) {
        return "internal node";
    } else {
        return "leaf";
    }

}

int main() {
    cin >> N;
    nodes = new Node [N];

    for (int k = 0; k < N; ++k) {
        nodes[k].left = nodes[k].right = nodes[k].parent = nil;
    }

    for (int i = 0; i < N; ++i) {
        int d;
        cin >> nodes[i].id >>  d;
        int l;
        for (int j = 0; j < d; ++j) {
            int c;
            cin >> c;
            if (j == 0) nodes[i].left = c;
            else nodes[l].right = c;

            l = c;
            nodes[c].parent = i;
        }
    }

    for (int m = 0; m < N; ++m) {
        cout << "node " << m << ": parent = " << nodes[m].parent << ", depth = " << depth(m) << ", " << type(m);
        int i = nodes[m].left;
        if (nodes[m].left == nil) {
            cout << ", []" << endl;
            continue; 
        }
        cout << ", [";
        while(true) {
            cout << i;
            if (nodes[i].right != nil) {
                cout << ", ";
                i = nodes[i].right;
            } else break;
        }
        cout << "]" << endl;
    }
    return 0;
}