p02279 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).

<image>
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.

<image>
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. 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.

Examples

Input

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


Output

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, []


Input

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


Output

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, []
```python
import sys

n = int(input())

nodes = [[-1, None, None] for _ in range(n)]

depths = [None for _ in range(n)]

for _ in range(n):
    nid, k, *children = list(map(int, sys.stdin.readline().split()))
    if k > 0:
        nodes[nid][1] = children[0]
    children.append(None)
    for i in range(k):
        cid = children[i]
        nodes[cid][0] = nid
        nodes[cid][2] = children[i+1]

def get_depth(nid):
    if depths[nid] is not None:
        return depths[nid]
    elif nodes[nid][0] == -1:
        depths[nid] = 0
        return 0
    else:
        depths[nid] = get_depth(nodes[nid][0]) + 1
        return depths[nid]
    

def get_children(nid):
    children = []
    cid = nodes[nid][1]
    while cid is not None:
        children.append(cid)
        cid = nodes[cid][2]
    return children

for nid in range(n):
    pid = nodes[nid][0]
    children = get_children(nid)
    depth = get_depth(nid)
    n_type = 'root' if pid == -1 else 'leaf' if len(children) == 0 else 'internal node'

    print('node {}: parent = {}, depth = {}, {}, {}'.format(nid, pid, depth, n_type, children))

```