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: Accepted)

#include <stdio.h>
#define MAX 100005
#define NIL -1

typedef struct node{
  int p; 
  int l;
  int r;
}Node;

void print(int u);
int rec(int u, int p);

Node T[MAX];
int n = 0, D[MAX];

int main(){
  int i = 0, j = 0;
  int d = 0, v = 0, c = 0, l = 0, r = 0;

  scanf("%d",&n);
  for(i = 0; i < n; i++){
    T[i].p = T[i].l = T[i].r = NIL;
  }

  for(i = 0; i < n; i++){
    scanf("%d%d",&v,&d);
    for(j = 0; j < d; j++){
      scanf("%d",&c);
      if(j == 0)T[v].l = c;
      else T[l].r = c;
      l = c;
      T[c].p = v;
    }
  }
  for(i = 0; i < n; i++){
    if(T[i].p == NIL)r = i;
  }

  rec(r, 0);

  for(i = 0; i < n; i++){
    print(i);
  }
  
  return 0;
}

void print(int u){
  int i = 0, c = 0;
  
  printf("node %d: ",u);
  printf("parent = %d, ",T[u].p);
  printf("depth = %d, ",D[u]);

  if(T[u].p == NIL)printf("root, [");
  else if(T[u].l == NIL)printf("leaf, [");
  else printf("internal node, [");

  c = T[u].l;
  for(i = 0; c != NIL; i++){
    if(i)printf(", ");
    printf("%d",c);
    c = T[c].r;
  }

  printf("]\n");

}

int rec(int u, int p){
  D[u] = p;
  if(T[u].r != NIL)rec(T[u].r, p);
  if(T[u].l != NIL)rec(T[u].l, p+1);
}
/*Input
  scanf("%d",&n);
  for(i = 0; i < n; i++){
    scanf("%d%d",&id,&k);
    T[i].sibling = k;
    if(k == 0){
      T[id].left[0] = -1;  子がいなければleftに-1  
    }
    T[i].parent = -2; もしrootだった時に-2のままのものを探す
    for(j = 0; j < k; j++){
      scanf("%d",&T[id].left[j]);
      T[T[id].left[j]].parent = id;
    }
  }

  Output
  for(i = 0; i < n; i++){
    printf("node %d: parent = %d, ",i,T[i].parent);
    
    depth = 0;
    j = i;
    while(T[j].parent != -1){
      j = T[j].parent;
      depth++;
    }
    printf("depth = %d, ",depth);
    
    if(T[i].parent == -2){
      printf("root, [");
    }
    else if(T[i].left[0] == -1)printf("leaf, []\n");
    else printf("internal node, [");

    if(T[i].left[0] != -1){
      for(l = 0; l < T[i].sibling; l++){
	if(l == T[i].sibling -1){
	  printf("%d]\n",T[i].left[l]);
	}
	else printf("%d, ",T[i].left[l]);
      }
    }
  }
*/