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>
#define MAX 100000
#define NIL -1
typedef struct{
  int p;
  int l;
  int r;
}Node;

Node T[MAX];
int n,D[MAX];//深さを代入

void print(int node_num){
    int i,c,flag=1;
    printf("node %d: ",node_num);//引数の番号
    printf("parent = %d, ",T[node_num].p);//引数ノードの親
    printf("depth = %d, ",D[node_num]);//引数ノードからrootまでの深さ
    if(T[node_num].p==(-1)) printf("root, ");
    else if(T[node_num].l==NIL) printf("leaf, ");//lに子の一番左のノードを入れる
    else printf("internal node, ");
    printf("[");
    for(c=T[node_num].l;c!=NIL;flag++){
        printf("%d",c);
        c=T[c].r;
        if(c!=NIL){
          printf(", ");
        }
      }
      printf("]");
      printf("\n");
}
//深さ探索(再起)
int rec(int node_num,int p){
  D[node_num]=p;//初期pは0値を逐一更新
  //初期化時全てのノードのp,r,lはNILになってる
  if(T[node_num].r!=NIL)rec(T[node_num].r,p);//最深部を探すため兄弟のノードからも子がないか調べる.
  if(T[node_num].l!=NIL)rec(T[node_num].l,p+1);//子のノードがあればpをカウントしてnode_numを子に変更
}

int main(){
  int i,j,depth,left,right,child,d,node_num;//dは次数
  scanf("%d",&n);
    for(i=0;i<n;i++){
      T[i].l=T[i].r=T[i].p=NIL;
    }
    for(i=0;i<n;i++){//ノードの個数
      scanf("%d",&node_num);
      scanf("%d",&d);//次数＝子の数
      for(j=0;j<d;j++){  //子入力
        scanf("%d",&child);
        if(j==0) T[node_num].l=child;//一回めに左端の子を代入
        else T[left].r=child;//以降の子は兄弟なのでrに代入
        left=child;//leftにchildのノードを入れておく
        T[child].p=node_num;//入れたノードの親を設定
   //ここまで子の入力
  }
}
    for(i=0;i<n;i++){
      if(T[i].p==NIL)right=i;
    }
    rec(right,0);
for(i=0;i<n;i++) print(i);
return 0;
}