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>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>

//============================================================================
#define pred(x) ((x)-1)
#define succ(x) ((x)+1)
//============================================================================
struct Node{
    long   id;
    long   item;
    struct Node *parent;
    struct Node **items;
};
typedef struct Node TNode;
//============================================================================
struct Chain{
    struct Chain *prev;
    TNode *node;
    struct Chain *next;
};
typedef struct Chain TChain;
TChain RootChain;
//============================================================================
void pushchain(TNode *node)
{
    TChain *chain;

    chain=&RootChain;
    while(chain->next!=NULL)
        chain=chain->next;
    chain->next=malloc(sizeof(TChain));
    chain->next->prev=chain;
    chain->next->next=NULL;
    chain->next->node=node;
}
//----------------------------------------------------------------------------
TNode *popchain(long id)
{
    TChain *chain; TNode *node;

    chain=&RootChain;
    while(chain->next!=NULL){
        chain=chain->next;
        if (chain->node->id==id){
            chain->prev->next=chain->next;
            if (chain->next!=NULL)
                chain->next->prev=chain->prev;
            node=chain->node;
            free(chain);
            return node;
        }
    }
    return NULL;
}
//----------------------------------------------------------------------------
TNode *search(TNode *node, long id)
{
    short i; TNode *ser;

    if (node->id==id)
        return node;
    if (!node->item)
        return NULL;
    for (i=0; i<node->item; i++)
        if ((ser=search(node->items[i], id))!=NULL)
            return ser;
    return NULL;
}
//----------------------------------------------------------------------------
//----------------------------------------------------------------------------
TNode *nodealloc(void)
{
    TNode *node;
    node=malloc(sizeof(TNode));
    if (node==NULL){
        fprintf(stderr, "not enough memry\n");
        exit(1);
    }
    memset(node, 0, sizeof(TNode));
    return node;
}
//----------------------------------------------------------------------------
void getnode(FILE *file, TNode **node)
{
    long i,n,id; TNode *ser, *pop;

    fscanf(file, "%ld", &id);
    fscanf(file, "%ld", &n);
    if (*node==NULL){
        *node=nodealloc();
        ser=*node;
    }else
        ser=search(*node, id);
    if (ser==NULL){
        ser=nodealloc();
        pushchain(ser);
    }
    ser->id=id;
    ser->item=n;
    if (ser->item){
        ser->items = malloc(sizeof(TNode*)*ser->item);
        memset(ser->items, 0, sizeof(TNode*)*ser->item);
        for (i=0; i<ser->item; i++){
            fscanf(file, "%ld", &id);
            if ((*node)->id==id){
                (*node)->parent=ser;
                ser->items[i]=*node;
                *node=ser;

            }else
            if ((pop=popchain(id))!=NULL){
                ser->items[i]=pop;
            }else{
                ser->items[i]=nodealloc();
                ser->items[i]->id=id;
            }
            ser->items[i]->parent=ser;
        }
    }
}
//----------------------------------------------------------------------------
short depth(TNode *node)
{
    short n;

    for (n=0; node->parent!=NULL; n++)
        node=node->parent;
    return n;
}
//----------------------------------------------------------------------------
char *root="root";
char *internalroot="internal node";
char *leaf="leaf";
char *nodetype(TNode *node)
{
    if (node->parent==NULL)
        return root;
    if (node->item)
        return internalroot;
    return leaf;
}
//----------------------------------------------------------------------------
void output(TNode *root, long n)
{
    long i,j,cnt,id; TNode *node;

    for (i=cnt=0; cnt<n; i++){
        node=search(root, i);
        if (node!=NULL){
            cnt++;
            id=(node->parent==NULL)?-1:node->parent->id;
            printf("node %ld: parent = %ld, depth = %d, %s, [",
                node->id, id, depth(node), nodetype(node));
            for (j=0; j<node->item; j++){
                if (j) printf(" ");
                printf("%d", node->items[j]->id);
                if (j+1!=node->item) printf(",");
            }
            printf("]\n");
        }
    }
}
//----------------------------------------------------------------------------
void input(FILE *file)
{
    long i,n,m; TNode *root, node; char buffer[20];

    memset(&RootChain, 0, sizeof(TChain));
    memset(&node, 0, sizeof(TNode));
    fgets(buffer, sizeof(buffer)-1, file);
    sscanf(buffer, "%ld", &n);
    for (i=0, root=NULL; i<n; i++){
        getnode(file, &root);
    }
    root->parent=NULL;
    output(root, n);
}
//============================================================================
#ifndef DESKTOP
int main()
{
    input(stdin);
    return 0;
}
#endif