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?

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

class Node(object):
    def __init__(self):
        self.id = None
        self.parent = -1
        self.children = None
        self.depth = -1

if __name__ == "__main__":
    n = int(input())
    Nodes = [Node() for _ in range(n)]
    for _ in range(n):
        inputs = list(map(int,input().split()))
        id = inputs[0]
        k = inputs[1]
        chilren = inputs[2:]
        Nodes[id].id = id
        Nodes[id].children = chilren
        for child in chilren:
            Nodes[child].parent = id
    
    root_id = -1
    for node in Nodes:
        if node.parent == -1:
            root_id = node.id
            node.depth = 0

    node_stack = [Nodes[root_id]]
    visit_node_id = {}
    out_str = [""]*n
    while len(node_stack) > 0 or len(visit_node_id) != n:
        node = node_stack.pop()
        visit_node_id[node.id] = 1

        id = node.id
        parent = node.parent
        depth = node.depth
        node_type = ""
        if parent == -1:
            node_type = "root"
        elif node.parent > -1 and len(node.children)>0:
            node_type = "internal node"
        elif node.parent > -1 and len(node.children) == 0:
            node_type = "leaf"
        children = node.children
        out_str[node.id] = "node "+str(id)+": parent = "+str(parent)+", depth = "+str(depth)+", "+node_type+", "+str(children)

        for child in node.children:
            if Nodes[child].id not in visit_node_id:
                Nodes[child].depth = node.depth+1
                node_stack.append(Nodes[child])
    
    for answer in out_str:
        print(answer)