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: Accepted)

#include<stdio.h>

typedef struct{
  int left;
  int right;
  int parent;
}node;

int main(){
  node T[100000];
  int d,n,i,j,k,id,num,c,cnt,p;

  scanf("%d",&n);
 for(i=0;i<n;i++){
   T[i].right = T[i].left = T[i].parent = -1;
 }
  for(i=0;i<n;i++){
    scanf("%d%d",&id,&k);
    
    for(j=0;j<k;j++){
      scanf("%d",&c);
      T[c].parent = id;
      if(j == 0) T[id].left = c;
      else T[p].right = c;
      p = c;
    }
  }
  for(i=0;i<n;i++){
    cnt = 0;
    p = T[i].parent;
    // printf("%d\n",p);
    while(p != -1){
      p = T[p].parent;
  cnt++;
    }
    
      printf("node %d: parent = %d, depth = %d, ",i,T[i].parent,cnt);

      if(T[i].parent == -1)printf("root, [");
      else if(T[i].left == -1)printf("leaf, [");
      else printf("internal node, [");

      p = T[i].left;
      if(p != -1){
	printf("%d",p);
	while(T[p].right != -1){
	  printf(", %d",T[p].right);
	  p = T[p].right;
	}
      }
      
      printf("]\n");
    
  }
    return 0;
  }



/*
根付き木

与えられた根付き木 T
の各節点 u

について、以下の情報を出力するプログラムを作成してください。

    u

の節点番号
u
の親の節点番号
u
の深さ
u
の節点の種類（根、内部節点または葉）
u

    の子のリスト

ここでは、与えられる木は n
個の節点を持ち、それぞれ 0 から n−1

の番号が割り当てられているものとします。
入力

入力の最初の行に、節点の個数 n
が与えられます。続く n

行目に、各節点の情報が次の形式で１行に与えられます。

id
k c1 c2 ... ck

id
は節点の番号、k は次数を表します。c1 c2 ...ck は 1 番目の子の節点番号、... k

番目の子の節点番号を示します。
出力

次の形式で節点の情報を出力してください。節点の情報はその番号が小さい順に出力してください。

node id
: parent = p, depth = d, type, [c1...ck

]

p
は親の番号を示します。ただし、親を持たない場合は -1 とします。d

は節点の深さを示します。

type

は根、内部節点、葉をそれぞれあらわす root、internal node、leaf の文字列のいずれかです。ただし、根が葉や内部節点の条件に該当する場合は root とします。

c1
...ck

は子のリストです。順序木とみなし入力された順に出力してください。カンマ空白区切りに注意してください。出力例にて出力形式を確認してください。
制約

    1≤n≤100,000

節点の深さは 20 を超えない。
任意の２つの節点間には必ず経路が存在する。
*/