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++14 (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <list>
#include <math.h>
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#define rep(i,n) for(int i=0;i<(n);i++)
#define crep(i) for(char i='a';i<='z';i++)
typedef long long ll;
#define psortsecond(A,N) sort(A,A+N,[](const pii &a, const pii &b){return a.second<b.second;});
#define psortfirst(A,N) sort(A,A+N,[](const pii &a, const pii &b){return a.first<b.first;});
#define All(x) (x).begin(),(x).end()
using namespace std;
using pii  = pair<int,int>;
long long gcd(long long a, long long b){return (b == 0 ? a : gcd(b, a%b));}
long long lcm(long long a, long long b){return a*b/gcd(a,b);}
int ctoi(const char c){
  if('0' <= c && c <= '9') return (c-'0');
  return -1;
}
#define MOD 1000000007
#define EPS 10e-8

ll d[100007],p;

int  depth(ll k[],ll parent[],ll i){
    if(parent[i]==-1){
        return 0;
    }
    else if(d[i]!=0){
        return d[i];
    }
    else{
        p = depth(k,parent,parent[i])+1;
        d[i] = p; 
        return p;
    }
}

int main(){
    ll N,k[100007],t,x,parent[100007];
    rep(i,100007){
        parent[i]=-1;
    }
    cin >> N;
    vector<vector<ll>> root(100007);
    rep(i,N){
        cin >> x;
        cin >> k[x];
        rep(j,k[x]){
            cin >> t;
            parent[t] = x;
            root[x].push_back(t);
        }
    }
    rep(i,N){
        // 節点番号
        printf("node %d: ",i);

        // 親
        if(parent[i]==-1){
            printf("parent = -1, ");
        }
        else{
            cout << "parent = "  << parent[i] << ", " ;
        }

        // 深さ
        printf("depth = %d, ",depth(k,parent,i));

        // 種類
        if(parent[i]==-1){
            printf("root, ");
        }
        else if(k[i]>0){
            printf("internal node, ");
        }
        else{
            printf("leaf, ");
        }

        // 子
        cout << "[";
        rep(j,k[i]){
            cout << root[i][j];
            if(j!=k[i]-1){
                cout << ", ";
            }
        }
        cout << "]";

        cout << endl;
    }
}