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....
class
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Can you provide 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:
.
 Vertex:
    _id = -1
    _children = []
    _parent = -1
    _depth = 0
    _type = "root"

    def __init__(self, i, c):
        self._id = i
        self._children = c

    def getId(self):
        return self._id

    def getChildren(self):
        return self._children

    def getParent(self):
        return self._parent

    def getDepth(self):
        return self._depth

    def getType(self):
        return self._type

    def setParent(self, p):
        self._parent = p

    def setDepth(self, d):
        self._depth = d

    def setType(self, t):
        self._type = t

    def print(self):
        print("node {}: parent = {}, depth = {}, {}, {}".format(self._id, self._parent, self._depth, self._type, self._children))

n = int(input())
tree = [0 for i in range(n)]

# 節点
for i in range(n):
    u = list(map(int, input().split(' ')))
    tree[u[0]] = Vertex(u[0], u[2:u[1]+2])

# 親を設定
for w in tree:
    for c in w.getChildren():
        tree[c].setParent(w.getId())
        tree[c].setType("internal node")

# 葉と親を判定
for w in tree:
    if w.getChildren() == []:
        w.setType("leaf")
    if w.getParent() == -1:
        w.setType("root")

# DFS
def dfs(v, depth):
    v.setDepth(depth)
    for child in v.getChildren():
        dfs(tree[child], depth+1)

# 根を探す
root = tree[0]
for w in tree:
    if w.getType() == "root":
        root = w

dfs(root, 0)

# 回答を出力
for w in tree:
    w.print()