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 MAX_NODE 100000

typedef int Item;

typedef struct node{
    Item item;
    int depth;
    struct node *parent;
    struct node *child;
    struct node *next;
} Node;

typedef struct tree{
    Node *root;
} Tree;

int getId(Node *node){
    return node->item;
}

int getParent(Node *node){
    if(node->parent != NULL) return node->parent->item;
    else return -1;
}

Node* getRoot(Node *node){
    if(node->parent == NULL) return node;
    else return getRoot(node->parent);
}

void setDepth(Node *root){
    if(root->parent == NULL) root->depth = 0;
    else root->depth = root->parent->depth + 1;

    if(root->child != NULL){
        Node *next = root->child;
        while(next != NULL){
            setDepth(next);
            next = next->next;
        }
    }
}

int getDepth(Node *node){
    return node->depth;
}

void printNode(Node *node){
    printf("node %d: ", node->item);
    printf("parent = %d, ", getParent(node));
    printf("depth = %d, ", getDepth(node));
    if(node->child == NULL) {
        if(node->parent == NULL) printf("root, ");
        else printf("leaf, ");
        printf("[]");
    }
    else {
        if (node->parent == NULL) printf("root, ");
        else printf("internal node, ");
        
        printf("[");
        Node *child = node->child;
        while(child != NULL){
            if(child->next == NULL) printf("%d]", child->item);
            else printf("%d, ", child->item);
            child = child->next;
        }

    }
    printf("\n");
}

int main(void){
    int n;
    scanf("%d\n", &n);
    
    Node node[MAX_NODE];
    
    int i;
    for(i = 0; i < n; i++){
        int id, k, j;
        scanf("%d%*c%d%*c", &id, &k);           
        node[id].item = id;
        for(j = 0; j < k; j++){
            int child;
            scanf("%d%*c", &child);
            
            node[child].parent =  &node[id];
            
            if(node[id].child == NULL) node[id].child = &node[child];
            else {
                Node *tail = node[id].child;
                while(tail->next != NULL) tail = tail->next;
                tail->next = &node[child];
            }
        }
    }
    
    Node *root = getRoot(&node[0]);
    setDepth(root);
    for(i = 0; i < n; i++){
        printNode(&node[i]);
    }


    return 0;
}