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 C (judge verdict: Accepted)

#include <stdio.h>

int get_uint() {
    int n = 0;
    int c = getchar_unlocked();
    if(c < 48 || 57 < c) return c;
    while(47 < c && c < 58) {
        n = 10 * n + (c & 0xf);
        c = getchar_unlocked();
    }
    return n;
}
void put_uint(int n) {
    if(!n) {
        putchar_unlocked('0');
        return;
    }
    char buf[11];
    int i = 0;
    while(n) buf[i++] = (char)(n % 10 + '0'), n /= 10;
    while(i--)putchar_unlocked(buf[i]);
}
void put_str(char *str) { while(*str) putchar_unlocked(*str++); }

typedef struct Node { int parent, left, right, depth; } Node;
void tree_depth(Node *tree, int node, int depth) {
    tree[node].depth = depth;
    for(int i = tree[node].left; ~i; i = tree[i].right) {
        tree_depth(tree, i, depth + 1);
    }
}
int main(int argc, char **argv) {
    int n = get_uint();
    Node tree[n];
    for(int i = 0; i < n; ++i) tree[i].parent = tree[i].left = tree[i].right = -1;
    for(int i = 0; i < n; ++i) {
        int id = get_uint();
        int k = get_uint();
        for(int j = 0, d; j < k; ++j) {
            int c = get_uint();
            tree[c].parent = id;
            if(j) tree[d].right = c;
            else tree[id].left = c;
            d = c;
        }
    }
    for(int i = 0; i < n; ++i) {
        if(tree[i].parent == -1) {
            tree_depth(tree, i, 0);
            break;
        }
    }
    for(int i = 0; i < n; ++i) {
        put_str("node ");
        put_uint(i);
        put_str(": parent = ");
        if(~tree[i].parent) put_uint(tree[i].parent);
        else put_str("-1");
        put_str(", depth = ");
        put_uint(tree[i].depth);
        put_str(~tree[i].parent ? ~tree[i].left ? ", internal node, [" : ", leaf, [" : ", root, [");
        if(~tree[i].left) {
            int j = tree[i].left;
            put_uint(j);
            for(j = tree[j].right; ~j; j = tree[j].right) {
                put_str(", ");
                put_uint(j);
            }
        }
        put_str("]\n");
    }
    return 0;
}