human
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

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

Constraints

* 1 ≤ n ≤ 100000

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....
Using python3 can you solve the prior task?
 Write me a Python code to print the information of each node in the following format ordered by IDs:?
assistant
#Here is Python code to print the information of each node in the following format ordered by IDs:
.

import sys
sys.setrecursionlimit(10**7)

class Node:
    def __init__(self,parent,left,right):
        self.parent = parent
        self.left = left
        self.right = right

def get_depth(T:dict,u:int):
    d = 0
    while T[u] != -1:
        u = T[u].parent
        d += 1

D = {}

def get_all_depth(T:dict,u:int,p:int):
    D[u] = p
    if T[u].right != -1:
        get_all_depth(T, T[u].right, p)
    if T[u].left != -1:
        get_all_depth(T,T[u].left,p+1)

def ret_children(T,u):
    children = []
    c = T[u].left
    while c != -1:
        children.append(c)
        c = T[c].right
    return children

def print_for_a_node(T:dict,node:int,D:dict):
    node_type = "leaf" if T[node].left == -1 else "internal node"
    if T[node].parent == -1:
        node_type = "root"
    parent = T[node].parent
    children = ret_children(T,node)
    print(f"node {node}: parent = {parent}, depth = {D[node]}, {node_type}, {children}")

n = int(input())

T = {k: Node(-1,-1,-1) for k in range(n)}

for _ in range(n):
    tmp = list(map(int,input().split()))
    if  tmp[1] == 0:
        continue
    T[tmp[0]].left = tmp[2]
    T[tmp[2]].parent = tmp[0]
    prev_sibling = tmp[2]
    for sib in tmp[3:]:
        T[prev_sibling].right = sib
        T[sib].parent = tmp[0]
        prev_sibling = sib

for node in range(n):
    if T[node].parent == -1:
        get_all_depth(T,node,0)
        break

for node in range(n):
    print_for_a_node(T,node,D)