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: Time Limit Exceeded)

#include <stdio.h>
#define N 100000

typedef struct{
  int id;
  int k;
  int c[20];
  int p;
  int d;
  int type;
  int flag;
}Node;
Node node[N];

int main(){
  int i,j,k,n,tmp;
  Node copy;
  
  scanf("%d",&n);
  if(n<0 || n>100001){
    printf("Error\n");
  }
  
  for(i=0;i<n;i++){
    scanf("%d",&node[i].id);
    scanf("%d",&node[node[i].id].k);
    for(j=0;j<node[node[i].id].k;j++){
      scanf("%d",&node[node[i].id].c[j]);
      node[node[node[i].id].c[j]].p = node[i].id;
      node[node[node[i].id].c[j]].flag = 1;
    }
    node[node[i].id].d = 0;                         /*depthの初期化*/
  }

  /*Search depth*/
  for(i=0;i<n;i++){
    if(node[node[i].id].k == 0) node[node[i].id].type = 2;   /*leaf*/
    else  node[node[i].id].type = 1;                /*internal node*/
    if(node[node[i].id].flag != 1){
      node[node[i].id].p =  -1;
      node[node[i].id].type = -1;                   /*root*/
    }
    tmp = node[i].p;
    /*  printf("node[i].p: %d,tmp: %d\n",node[i].p,tmp);
     printf("node[tmp].p: %d\n",node[1].p);
     j=0;*/
    if(tmp == -1)continue;
    while(tmp != -1){
      //printf("j:%d\n",j);
      node[i].d++;
      tmp = node[tmp].p;
      // printf("tmp: %d\n",tmp);
      // j++;
      if(j>3)break;
    }
  }
   
  /*Input*/
   for(i=0;i<n;i++){
     printf("node %d: parent = %d, depth = %d, ",i,node[i].p,node[i].d);
     if(node[i].type == -1) printf("root, [");             
     else if(node[i].type == 1) printf("internal node, [");
     else printf("leaf, [");                               
     for(j=0;j<node[i].k;j++){
       if(j == 0){
	 printf("%d",node[i].c[0]);
	 continue;
       }
       if(j < node[i].k){
	 printf(", %d",node[i].c[j]);
       }
     }
     printf("]\n");
   }
   
   return 0;
}