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 MAX 100001

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

Tree tree[MAX];
int n,D[MAX];

void count(int,int);

int main(){
  int i,j,d,v,c,left,right;

  scanf("%d",&n);

  for( i = 0 ; i < n ; i++ )
    {
      tree[i].parent = -1;
      tree[i].right = -1;
      tree[i].left = -1;
    }

  for( i = 0 ; i < n ; i++ )
    {
      scanf("%d%d",&v,&d);
      for( j = 0 ; j < d ; j++ )
	{
	  if( j == 0 ) tree[v].left=c;
	  else tree[left].right = c;
	  left = c;
	  tree[c].parent = v;
	}
    }

  for( i = 0 ; i < n ; i++ ){
    if(tree[i].parent == -1) right =i;
  }

  count(right,0);

  for(i = 0 ; i < n ; i++ ){
printf("node %d: parent = %d, depth = %d, ",i,tree[i].parent,D[i]);
 if(tree[i].parent == -1 ) printf("root, [");
 else if(tree[i].left == -1 ) printf("leaf, [");
 else printf("internal node, [");  

 for(j = 0,c=tree[i].left;c!=-1;j++,c=tree[c].right){
   if(j) printf(", ");
   printf("%d",c);
 }
 printf("]\n");
  }

  return 0;
}

void count(int u,int p){
  D[u]=p;
  if(tree[u].right!=-1) count(tree[u].right,p);
  if(tree[u].left!=-1) count(tree[u].left,p+1);
}