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 NIL -1

int depth(int);

struct Node{
  int parent;
  int left;
  int right;
};

struct Node T[100000];

int main(){

  int n,id,k,c[100000],i,j,m,de,count=0;

  scanf("%d",&n);

  for(i=0;i<n;i++){
    T[i].right=NIL;
  }

  while(1){

    scanf("%d%d",&id,&k);
    count++;

    
    // printf("wa-i1\n");

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

    
    if(count==1){
      
      T[id].parent=NIL;
      T[id].left=c[0];
      T[id].right=NIL;
      //printf("T:%d c[0]:%d\n",T[id].left,c[0]);
    } 
    
    if(k!=0){
      
      T[id].left=c[0];
      
      for(j=0;j<k;j++){
	T[c[j]].parent=id;
	
	if(j==k-1){T[c[j]].right=NIL;} 
	else {T[c[j]].right=c[j+1];}
	
      }
    }
    
    else if(k==0){
      
      T[id].left=NIL;
      
      
      // T[id].right=NIL;
	 }

    if(count==n)break;
  }

  // printf("wa-i2\n");
  
  for(i=0;i<n;i++){

    // printf("wa-i4\n");
    
    de = depth(i);

    //printf("wa-i3\n");
    
    printf("node %d: parent = %d, depth = %d,",i,T[i].parent,de);
    
    if(T[i].parent==NIL){printf(" root, ");}
    else if(T[i].left==NIL){printf(" leaf, ");}
    else {printf(" internal node, ");}
    
    printf("[");

    j=T[i].left;

    //printf("\nj:%d\n",j);
    
    while(j!=NIL){

      if(T[i].left==j){
	printf("%d",j);
       
      }
      else{
	printf(", %d",j);
      }
      j=T[j].right;
    }
    
    printf("]\n");
  }

  //printf("node1:right:%d\n",T[].right);


  
  return 0;
}

int depth(int num){

  if(T[num].parent==NIL)return 0;

  return depth(T[num].parent)+1;
  
}