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 JAVA (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigDecimal;
import java.util.ArrayList;
/**
 * 境界部分には気をつける(そこだけ別で処理するというのもあり)
 * 逆再生することで簡単になることもある
 */
public class Main {
	public int cnt = 0;
	public static final BigDecimal TWO = new BigDecimal(2);
	public static final BigDecimal THREE = new BigDecimal(3);

    public Main() {
    }

    public static void main(String[] args ) {
            Main main = new Main();
            main.exec();
    }

    public void exec() {
        BufferedReader bf = null;
        try{
            bf = new BufferedReader(new InputStreamReader(System.in));
            exec_body(bf);
        } catch (IOException e) {
            e.printStackTrace();
        }
        finally{
            if( bf != null ) {
                try {
                    bf.close();
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        }
    }

    public class Node {
    		int num = 0;
    		Node parent = null;
    		ArrayList<Node> nodes = new ArrayList<Node>();
    }

    public void exec_body( BufferedReader bf ) throws IOException {
    		String line = bf.readLine();
    		int nodeLength = Integer.parseInt(line);
    		Node[] nodes = new Node[nodeLength];
    		for( int ii = 0; ii < nodeLength; ii++ ) {
    			Node node = new Node();
    			node.num = ii;
    			nodes[ii] = node;
    		}

    		int lineCnt = 0;
    		do {
    			line = bf.readLine();
    			String[] lines = line.split(" ");
    			if( !"0".equals(lines[1]) ) {
        			int parentNum = Integer.parseInt(lines[0]);
        			Node parent = nodes[parentNum];
        			for( int ii = 2; ii < lines.length; ii++ ) {
        				int childNum = Integer.parseInt(lines[ii]);
        				Node child = nodes[childNum];
        				child.parent = parent;
        				parent.nodes.add(child);
        			}
    			}
    			lineCnt++;
    		} while( lineCnt != nodeLength );

    		for( int ii = 0; ii < nodes.length; ii++ ) {
    			// node num
    			Node node = nodes[ii];
    			StringBuilder bld = new StringBuilder();
    			bld.append("node ");
    			bld.append(node.num);

    			// parent
    			bld.append(": parent = ");
    			int parentNum = ( node.parent == null ) ? -1 : node.parent.num;
    			bld.append(parentNum);

    			// depth
    			bld.append(", depth = ");
    			bld.append(getDepth(node));
    			bld.append(", ");

    			// type
    			if( parentNum == -1 ) {
    				bld.append("root, ");
    			} else if( node.nodes.isEmpty() ) {
    				bld.append("leaf, ");
    			} else {
    				bld.append("interna node, ");
    			}

    			// childs
    			bld.append("[");
    			for( int kk = 0; kk < node.nodes.size(); kk++ ) {
    				if( kk != 0) {
    					bld.append(", ");
    				}
    				Node childNode = node.nodes.get(kk);
    				bld.append(childNode.num);
    			}
    			bld.append("]");
    			System.out.println(bld.toString());
    		}
    }

    int getDepth( Node node ) {
    		int depth = 0;
    		while( node.parent != null ) {
    			node = node.parent;
    			depth++;
    		}

    		return depth;
    }

    public int max( int[] ary ) {
    		int ret = Integer.MIN_VALUE;
    		for( int ii = 0; ii < ary.length; ii++ ) {
    			if( ret < ary[ii] ) {
    				ret = ary[ii];
    			}
    		}

    		return ret;
    }

    public int min( int[] ary ) {
		int ret = Integer.MAX_VALUE;
		for( int ii = 0; ii < ary.length; ii++ ) {
			if( ret > ary[ii] ) {
				ret = ary[ii];
			}
		}

		return ret;
    }

    public static int[] toIList( String str ) {
        String[] elements = str.split(" ");
        int[] ret = new int [elements.length];
        for( int i = 0; i < elements.length; i++ ) {
            ret[i] = Integer.valueOf(elements[i]);
        }

        return ret;
    }

    public static void print( int[] ary ) {
    		if( ary == null || ary.length == 0 ) {
    			return;
    		}

    		StringBuilder bd = new StringBuilder();
    		for( int i = 0; i < ary.length; i++ ) {
    			if( i != 0 ) {
    				bd.append(" ");
    			}
    			bd.append(ary[i]);
    		}

    		System.out.print(bd.toString());
    }
}