Problem:
Rooted Trees

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

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. If the node does not have a parent, print -1.

d is depth of the node.

type is a type of nodes represented by a string (root, internal node or leaf). If the root can be considered as a leaf or an internal node, print root.

c1...ck is the list of children as a ordered tree.

Please follow the format presented in a sample output below.

Constraints

1 ≤ n ≤ 100000

Sample Input 1

13
0 3 1 4 10
1 2 2 3
2 0
3 0
4 3 5 6 7
5 0
6 0
7 2 8 9
8 0
9 0
10 2 11 12
11 0
12 0

Sample Output 1

node 0: parent = -1, depth = 0, root, [1, 4, 10]
node 1: parent = 0, depth = 1, internal node, [2, 3]
node 2: parent = 1, depth = 2, leaf, []
node 3: parent = 1, depth = 2, leaf, []
node 4: parent = 0, depth = 1, internal node, [5, 6, 7]
node 5: parent = 4, depth = 2, leaf, []
node 6: parent = 4, depth = 2, leaf, []
node 7: parent = 4, depth = 2, internal node, [8, 9]
node 8: parent = 7, depth = 3, leaf, []
node 9: parent = 7, depth = 3, leaf, []
node 10: parent = 0, depth = 1, internal node, [11, 12]
node 11: parent = 10, depth = 2, leaf, []
node 12: parent = 10, depth = 2, leaf, []

Sample Input 2

4
1 3 3 2 0
0 0
3 0
2 0

Sample Output 2

node 0: parent = 1, depth = 1, leaf, []
node 1: parent = -1, depth = 0, root, [3, 2, 0]
node 2: parent = 1, depth = 1, leaf, []
node 3: parent = 1, depth = 1, leaf, []

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

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in Python3 (judge verdict: Runtime Error)

NIL = -1


class Node:
    parent = NIL
    left = NIL
    right = NIL
    label = ""


def show_info(u):
    global T, D
    print('node {0}: parent = {1}, depth = {2}, {3}, {4}'.format(u, T[u].parent, D[u], T[u].label, getChildren(u)))


def setLabel(n):
    for i in range(n):
        if T[i].parent == NIL:
            T[i].label = "root"
        elif T[i].left == NIL:
            T[i].label = "leaf"
        else:
            T[i].label = "internal node"


def getChildren(u):
    global T
    c = T[u].left
    R =[]
    while c != NIL:
        R.append(c)
        c = T[c].right
    return R


def setDepth(u,p):
    global T, D
    D[u] = p
    if T[u].right != NIL:
        setDepth(T[u].right, p)
    if T[u].left != NIL:
        setDepth(T[u].left, p + 1)


if __name__ == "__main__":
    n = int(input())
    T = [None] * n
    D = [0] * n
    for i in range(n):
        T[i] = Node()
    for i in range(n):
        tmp = list(map(int, input().split()))
        id = tmp.pop(0)
        k = tmp.pop(0)
        C = tmp
        if k != 0:
            T[id].left = C[0]
            for j in range(len(C)):
                T[C[j]].parent = id
            for j in range(len(C) - 1):
                T[C[j]].right = C[j + 1]
    setLabel(n)
    for i in range(n):
        if T[i].label == "root":
            root = i
    # ??±????¨????
    setDepth(root, 0)
    for i in range(n):
        show_info(i)