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 program to print the information of each node in the following format ordered by IDs:
.

from collections import deque
import sys
sys.setrecursionlimit(100000)

class Node():
    def __init__(self, p=None, l=None, r=None):
        self.parent = p
        self.left = l
        self.right = r
        self.depth = None

n = int(input())
nodes = [Node() for i in range(n)]
for i in range(n):
    id, k, *c = list(map(int, input().split(' ')))
    if k == 0: continue
    nodes[id].left = c[0]
    nodes[c[0]].parent = id
    for j in range(1, k):
        nodes[c[j-1]].right = c[j]
        nodes[c[j]].parent = id

root = None
for i in range(n):
    if nodes[i].parent == None:
        root = i

def get_childs(i):
    if nodes[i].left == None: return []
    j = 0
    childs = [nodes[i].left]
    while nodes[childs[j]].right != None:
        childs.append(nodes[childs[j]].right)
        j += 1
    return childs

# 再帰だと segmentation faultが発生する
# while で実装

def set_depth(s):
    nodes[s].depth = 0
    queue = deque([s])
    while len(queue) > 0:
        u = queue.popleft()
        par_depth = nodes[u].depth
        next_child = nodes[u].left
        while next_child != None:
            nodes[next_child].depth = par_depth + 1
            queue.append(next_child)
            next_child = nodes[next_child].right

set_depth(root)

for i in range(n):
    nodetype = 'internal node'
    childs = get_childs(i)
    if childs == []:
        nodetype = 'leaf'
    parent = nodes[i].parent
    if parent == None:
        parent = -1
        nodetype = 'root'
    print('node ' + str(i) + ': parent = ' + str(parent) + ', depth = ' + str(nodes[i].depth) + ', ' + nodetype + ', ' + str(childs))