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++11 (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <random>
using namespace std;

#define REP(i,n) for(int (i)=0; (i)<(n); (i)++)
#define For(i,a,b) for(int (i)=(a); (i)<(b); (i)++)
#define ALL(v) v.begin(), v.end()

class Tree {
    int n;
    vector<vector<int>> adjlist;
    vector<vector<int>> inv_adjlist;

public:
    Tree(int num);
    int getParent(int id);
    void addNode(int parentId, int childId);
    int getDepth(int id);
    string getType(int id);
    vector<int> getChild(int id);
};

Tree::Tree(int num) {
    n = num;
    REP(i,num) {
        vector<int> v;
        adjlist.push_back(v);
        inv_adjlist.push_back(v);
    }
}

void Tree::addNode(int parentId, int childId) {
    adjlist[parentId].push_back(childId);
    inv_adjlist[childId].push_back(parentId);
}

int Tree::getParent(int id) {
    REP(i,n) {
        if (find(ALL(adjlist[i]), id) != adjlist[i].end()) {
            return i;
        }
    }
    return -1;
}

int Tree::getDepth(int id) {
    int depth = 0;
    vector<int> parent = inv_adjlist[id];
    while(parent.size()) {
        parent = inv_adjlist[parent[0]];
        depth++;
    }
    
    /* slow code
    int parent = id;
    while (1) {
        parent = getParent(parent);
        if (parent != -1) {
            depth++;
        }else {
            break;
        }
    }*/
    return depth;
}

string Tree::getType(int id) {
    if (getParent(id) == -1) {
        return "root";
    }else if (adjlist[id].empty()) {
        return "leaf";
    }else {
        return "internal node";
    }
    return "";
}

vector<int> Tree::getChild(int id) {
    return adjlist[id];
}

int main() {
    int n;
    cin >> n;
    Tree tree = Tree(n);
    REP(i,n) {
        int id, deg;
        cin >> id >> deg;
        int k;
        REP(j,deg) {
            cin >> k;
            tree.addNode(id, k);
        }
    }
    
    REP(i,n) {
        vector<int> children = tree.getChild(i);
        cout << "node " << i << ": parent = ";
        cout << tree.getParent(i) << ", depth = ";
        cout << tree.getDepth(i) << ", ";
        cout << tree.getType(i) << ", [";
        REP(j,children.size()) {
            cout << children[j];
            if (j < children.size()-1) {
                cout << ", ";
            }
        }
        cout << "]" << endl;
    }
    return 0;
}