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>
#include <string.h>

#define N 100000
typedef struct{
  int par, left_child, right_sibling, depth;
  char status[15];
}Node;

Node tree[N];
int n, nchild[N];
char root[N];

void calc_depth(int p, int d){
  tree[p].depth = d;
  if(tree[p].right_sibling != -1) calc_depth(tree[p].right_sibling, d);
  if(tree[p].left_child != -1) calc_depth(tree[p].left_child, d+1);
}

void print_node(int p, int cnt, int size){
  if(cnt + 1 == size) printf("%d", p);
  else{
    printf("%d, ", p);
    print_node(tree[p].right_sibling, cnt + 1, size);
  }
}

main(){
  int i, j, r;
  scanf("%d", &n);
  memset(root, 1, n);
  for(i=0;i<n;i++) tree[i].left_child = tree[i].right_sibling = -1;
  for(i=0;i<n;i++){
    int idx;
    scanf("%d", &idx);
    scanf("%d", &nchild[idx]);
    if(nchild[idx] == 0) strcpy(tree[idx].status, "leaf");
    else strcpy(tree[idx].status, "internal node");
    for(j=0;j<nchild[idx];j++){
      int in, pre;
      scanf("%d", &in);
      tree[in].par = idx;
      root[in] = 0;
      if(j == 0) tree[idx].left_child = in;
      else tree[pre].right_sibling = in;
      pre = in;
    }
  }
  for(i=0;i<n;i++) if(root[i]) r = i;
  tree[r].par = -1;
  strcpy(tree[r].status, "root");
  calc_depth(r, 0);
  for(i=0;i<n;i++){
    printf("node %d: parent = %d, depth = %d, %s, [", i, tree[i].par, tree[i].depth, tree[i].status);
    if(tree[i].left_child != -1) print_node(tree[i].left_child, 0, nchild[i]);
    printf("]\n");
  }
  return 0;
}