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: Memory Limit Exceeded)

#include <stdio.h>
#define N 100000
 
typedef struct{
  int id,p,d,k;
  int jisu[1000];
}Node;
// tree{0, 1, 2, 3, 4, 5, 6}                                                                  
// tree                                                                                       
Node tree[N];
 
int jisu(int,int);
 
int kaisu = 0,n;
main(){
 
  int j,i,nodeb;
 
  scanf("%d",&n);
 
  for(j = 0 ; j < n; j++){
    tree[j].d = 0;
    tree[j].p = -2;
    tree[j].k = 0;
  }
 
  for(j = 0 ; j < n; j++){
    scanf("%d",&nodeb);
    // 入力されたノード番号のIDをtreeの”入力されたノード番号”番目に設定する                   
    tree[nodeb].id = nodeb;
    scanf("%d",&tree[nodeb].k);
    jisu(tree[nodeb].k,nodeb);
  }
 
  for(j = 0 ; j < n; j++ ){
    if(tree[j].p == -2)tree[j].p = -1;
     printf("node %d: parent = %d, depth = %d,",tree[j].id,tree[j].p,tree[j].d);
 
    if(tree[j].p == -1){
      printf(" root, [");
      for(i = 0 ; i < tree[j].k ; i++ ){
        printf("%d",tree[j].jisu[i]);
        if(i != tree[j].k - 1){
          printf(", ");
        }
 
      }
      printf("]");
    }
 
    //&& tree[j].id != 0                                                                      
 
    if(tree[j].k == 0 && tree[j].p != -1){
      printf(" leaf, []");
    }
 
    if(tree[j].k != 0 && tree[j].p != -1){
      printf(" internal node, [");
     for(i = 0 ; i < tree[j].k ; i++ ){
        printf("%d",tree[j].jisu[i]);
        if(i != tree[j].k - 1){
          printf(", ");
        }
     }
      printf("]");
    }
 
    printf("\n");
  }
 
  return 0;
 
}
 
 
 
 
int jisu(int k,int now){
  int ii,g;
 
    for(ii = 0 ; ii < k; ii++){
      scanf("%d",&tree[now].jisu[ii]);
      if(tree[now].jisu[ii] == 0 && tree[tree[now].jisu[ii]].d != 0){
        tree[now].p = -1;
      }
      //次数に親のidいれる                                                                    
      tree[tree[now].jisu[ii]].p = now;
      tree[tree[now].jisu[ii]].d = tree[now].d + 1;
 
    for(g = 0 ; g < n; g++){
      if(tree[g].p == -1 || tree[g].p == -2) continue;
      if(tree[g].d == tree[tree[g].p].d) tree[g].d++;
    }
 
    }
    return 1;
}