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>

struct Tree {
  int id;
  int k;
  int c[20];
  int parent;
  int depth;
};

int depth(int);
struct Tree data[100];

int main()
{
  int n, i, j, flag, k;
  struct Tree temp;
  char *str[] = {"root", "internal node", "leaf"};

/* ??\??? */
  scanf("%d", &n);

  for(i = 0; i < n; i++){
    scanf("%d%d", &data[i].id, &data[i].k);
    for(j = 0; j < data[i].k; j++){
      scanf("%d", &data[i].c[j]);
    }
  }

  data[0].parent = -1;
  data[0].depth = 0;

/* parent?????? */
  for(i = 0; i < n; i++){
    for(j = 0; j < data[i].k; j++){
      for(k = 0; k < n; k++){
	if(i == 0) data[i].parent = -1;
	if(data[i].c[j] == data[k].id){
	  data[k].parent = data[i].id;
	  data[k].depth = data[j].id;
	}
      }
    }
  }

/* depth?????? */
  /* for(i = 0; i < n; i++){ */
  /*   if(data[i].parent == -1) data[i].depth = 0; */
  /*   else if(data[i].parent == data[i-1].depth && data[n-1].depth == 0) data[i].depth = 1; */
  /*   else if(data[i].parent == data[i-1].depth && data[n-1].depth == 1) data[i].depth = 2; */

/* ?????????sort */
  flag = 1;
  j = 0;
  while(flag){
    if(n == 1) break;
    flag = 0;
    for(i = n-1; i >= 1; i--){
      if(data[i].id < data[i-1].id){
	temp = data[i-1];
	data[i-1] = data[i];
	data[i] = temp;
	flag = 1;
	j++;
      }
    }
  }

  /* depth */
  for(i = 0; i < n; i++){
    data[i].depth = depth(data[i].parent);
  }

/* ?????? */
  for(i = 0; i < n; i++){
    printf("node %d: ", data[i].id);
    printf("parent = %d, ", data[i].parent);
    printf("depth = %d, ", data[i].depth);
    
    if(data[i].parent == -1) printf("%s, ", str[0]);
    else if(data[i].parent != -1 && data[i].c[0] == 0) printf("%s, ", str[2]);
    else printf("%s, ", str[1]);

    printf("[");
    for(j = 0; j < data[i].k; j++){
      if(j == 0) printf("%d", data[i].c[j]);
      else printf(", %d", data[i].c[j]);
    }
    
    printf("]\n");
    /* printf("\n"); */
  } 
  /* parent = %d, depth = %d, %s, [1, 4, 10] */
  
  return 0;
}
  
int depth(int p)
{
  int count=0;

  if(p == -1) return 0;
  else{
    count++;
    return count + depth(data[p].parent);
  }
}