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?
assistant
#Here is Python code to print the information of each node in the following format ordered by IDs:
.

import sys
sys.setrecursionlimit(2 ** 20)

class Node():
    def __init__(self, parent=None, left=None, right=None):
        self.parent = parent
        self.left = left
        self.right = right

    def __repr__(self):
        return '<Node object: parent={}, left={}, right={}>' .format(self.parent, self.left, self.right)

def set_depth(u, p):
    D[u] = p
    if not T[u].right is None:
        set_depth(T[u].right, p)
    if not T[u].left is None:
        set_depth(T[u].left, p+1)

def get_children(u):
    res = []
    c = T[u].left
    while c is not None:
        res.append(c)
        c = T[c].right
    return res

def get_type(node):
    if node.parent is None:
        return 'root'
    elif node.left is None:
        return 'leaf'
    else:
        return 'internal node'

def get_parent(node):
    return -1 if node.parent is None else node.parent

n = int(input())
T = [Node() for i in range(n)]
for _ in range(n):
    line = list(map(int, input().split()))
    i, k = line[:2]
    if k:
        T[i].left = line[2]
        prev = None
        for c in line[2:]:
            T[c].parent = i
            if prev:
                T[prev].right = c
            prev = c

D = [0] * n
for i in range(n):
    if T[i].parent is None:
        set_depth(i, 0)
        break

for i in range(n):
    print('node {}: parent = {}, depth = {}, {}, {}'.format(
        i, get_parent(T[i]), D[i], get_type(T[i]), get_children(i)))