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>
#include<stdlib.h>

int main(){
  int i,j,n,id,l,s=0,t;
  int **node,*d,*p,*k;
  char **r;
  
  scanf("%d",&n);

  node = (int **)malloc(sizeof(int *) * n);
  d = (int *)malloc(sizeof(int) * n);
  p = (int *)malloc(sizeof(int) * n);
  k = (int *)malloc(sizeof(int) * n);
  r = (char **)malloc(sizeof(char *) * n);
  
  for(i = 0 ; i < n ; i++){

    scanf("%d",&id);
    scanf("%d",&k[id]);
    
    node[id] = (int *)malloc(sizeof(int) * k[id]);

    for(j = 0 ; j < k[id] ; j++){
      scanf("%d",&node[id][j]);
    }
  }

  for(i = 0 ; i < n ; i++){
    s = 0;
    if(k[i] > 0){
      for(j = 0 ; j < n ; j++){
	
	if(s == 1)break;
	
	for(l = 0 ; l < k[j] ; l++){
	  if(i == node[j][l]){
	    p[i] = j;

	    r[i] = (char *)malloc(sizeof(char) * 16);
	    r[i] = "internal node, ";
	    s = 1;
	    break;
	  }
	}
	
	if(s == 0){
	  p[i] = -1;
	  r[i] = (char *)malloc(sizeof(char) * 7);
	  r[i] = "root, ";
	}
      }
    }
    
    else if(k[i] == 0){
      for(j = 0 ; j < n ; j++){
	for(l = 0 ; l < k[j] ; l++){
	  if(i == node[j][l]){
	    p[i] = j;
	    r[i] = (char *)malloc(sizeof(char) * 7);
	    r[i] = "leaf, ";
	    break;
	  }
	}
      }
    }
    
  }

  

  for(i = 0 ; i < n ; i++){
    s = p[i];
    d[i] = 0;
    while(s != -1){
      d[i]++;	    
      t = p[s];
      s = t;
    }
  }
    


  for(i = 0 ; i < n ; i++){
    printf("node %d: ",i);
    printf("parent = %d, ",p[i]);
    printf("depth = %d, ",d[i]);
    printf("%s",r[i]);
    
    printf("[");
    
    for(j = 0 ; j < k[i] ; j++){
      printf("%d",node[i][j]);
      if(j < k[i] - 1) printf(", ");
    }
    printf("]\n");
  }

  free(node);
  free(d);
  free(p);
  free(k);
  free(r);
  
  return 0;
}