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<bits/stdc++.h>
using namespace std;
typedef  long long ll;
using vll=vector<ll>;
using vi=vector<int>;
using vb=vector<bool>;
using pii=pair<int,int>;

template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b;return true;}return false;}
template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b;return true;}return false;}

//pow(llpow,modpow)
template<class T> ll llpow(ll x,T n){ll ans=1;if(x==0)ans=0;while(n){if(n&1)ans*=x;x*=x;n>>=1;}return ans;}
long long modpow(long long a, long long n, long long mod) {long long res = 1;while (n > 0) {if (n & 1) res = res * a % mod;a = a * a % mod;n >>= 1;}return res;}
//最大公約数
template<class T> inline T gcd(T x,T y){if(y==0)return x; else {return gcd(y,x%y);}}
//最小公倍数
template<class T> inline T lcm(T x,T y){return x/gcd(x,y)*y;}
//逆元
long long modinv(long long a, long long m) {long long b = m, u = 1, v = 0;while (b) {long long t = a / b;a -= t * b; swap(a, b);u -= t * v; swap(u, v);}u %= m;if (u < 0) u += m;return u;}

#define rep(i, begin_i, end_i) for (ll i = (ll)begin_i; i < (ll)end_i; i++)

long long INF = 1LL<<60;;
typedef struct node {
    int identifier;
    bool is_parent=false;
    int depth;
    int parent=-1;
    list<int> children;
}Node;

string cnvkind(bool p,bool r){
    if(p&&r)return "root";
    else if(p&&!r)return "internal node";
    else return "leaf";
}

void print(Node n,bool is_root){
    cout<<"node "<<n.identifier<<": parent = "<<n.parent<<", depth = "<<n.depth<<", "+cnvkind(n.is_parent,is_root)<<", [";
    int cnt=0;
    for(auto itr : n.children){
        if(cnt++)cout<<", ";
        cout<<itr;
    }
    cout<<"]"<<endl;
}

int main( ){
    int N;
    cin>>N;
    Node info[N];
    vb is_root(N,true);;
    rep(i,0,N){
        int id;
        cin>>id;
        info[id].identifier=id;
        int k;
        cin>>k;
        if(k||N==1)info[id].is_parent=true;
        rep(j,0,k){
            int c;
            cin>>c;
            info[id].children.push_back(c);
            is_root[c]=false;
            info[c].parent=id;
        }
    }
    int root,depth=0;
    rep(i,0,N)if(is_root[i])root=i;
    queue<pii> q;
    q.emplace(root,depth);
    while(!q.empty()){
        pii tmp=q.front(); q.pop();
        info[tmp.first].depth=tmp.second;
        for(int itr:info[tmp.first].children){
            q.emplace(itr,tmp.second+1);
        }
    }
    rep(i,0,N)print(info[i],is_root[i]);
    return 0;
}