Problem:
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Depth First Search

Depth-first search (DFS) follows the strategy to search ”deeper” in the graph whenever possible. In DFS, edges are recursively explored out of the most recently discovered vertex $v$ that still has unexplored edges leaving it. When all of $v$'s edges have been explored, the search ”backtracks” to explore edges leaving the vertex from which $v$ was discovered.

This process continues until all the vertices that are reachable from the original source vertex have been discovered. If any undiscovered vertices remain, then one of them is selected as a new source and the search is repeated from that source.

DFS timestamps each vertex as follows:

$d[v]$ records when $v$ is first discovered.

$f[v]$ records when the search finishes examining $v$’s adjacency list.

Write a program which reads a directed graph $G = (V, E)$ and demonstrates DFS on the graph based on the following rules:

$G$ is given in an adjacency-list. Vertices are identified by IDs $1, 2,... n$ respectively.

IDs in the adjacency list are arranged in ascending order.

The program should report the discover time and the finish time for each vertex.

When there are several candidates to visit during DFS, the algorithm should select the vertex with the smallest ID.

The timestamp starts with 1.

Input

In the first line, an integer $n$ denoting the number of vertices of $G$ is given. In the next $n$ lines, adjacency lists of $u$ are given in the following format:

$u$ $k$ $v_1$ $v_2$ ... $v_k$

$u$ is ID of the vertex and $k$ denotes its degree. $v_i$ are IDs of vertices adjacent to $u$.

Output

For each vertex, print $id$, $d$ and $f$ separated by a space character in a line. $id$ is ID of the vertex, $d$ and $f$ is the discover time and the finish time respectively. Print in order of vertex IDs.

Constraints

$1 \leq n \leq 100$

Sample Input 1

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

Sample Output 1

1 1 8
2 2 7
3 4 5
4 3 6

Sample Input 2

6
1 2 2 3
2 2 3 4
3 1 5
4 1 6
5 1 6
6 0

Sample Output 2

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

This is example for Sample Input 2 (discover/finish)

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)

using System;
using System.Windows;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static int time;

    static void Main(string[] args)
    {
        time = 0;
        //入力を受け取る
        var n = int.Parse(Console.ReadLine());
        mynode[] M = new mynode[n + 1];
        var S = new MyStack();

        for(int i = 1; i < n + 1; i++)
        {
            M[i] = new mynode();
        }

        //各ノードの情報を初期化する
        for (int i = 1; i < n + 1; i++)
        {
            var ukv = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
            var u = ukv[0];
            var k = ukv[1];
            M[i].initialize(u, k);

            for (int j = 0; j < k; j++)
            {
                M[i].RegistAdjacenxy(M[ukv[2 + j]]);
            }
        }


        //スタックで探索を管理
      　for(int i = 1; i < n + 1; i++){
        RegistStack(M[i], S);
       }
        //探索終了時間の登録
        //while (S.IsEmpty() != -1)
        //{
          //  time++;
           // S.pop().finish = time;
       // }

        //結果の出力
        for (int i = 1; i < n + 1; i++)
        {
            Console.WriteLine(M[i].id + " " + M[i].find + " " + M[i].finish);
        }

        
    }

    static void RegistStack(mynode m, MyStack S)
    {
        //対象ノードが登録済かどうかを調べる
        if (m.find == -1)
        {
            time++;
            m.find = time;
            foreach (mynode k in m.adjacency)
            {
                S.push(m);
                if (k != null)
                {
                    RegistStack(k, S);
                }
            }
            time++;
            m.finish = time;
          　　
        }
    }
}

class MyStack
{
    public mynode[] datas = new mynode[1000];
    public int point = -1;
    public void push(mynode data)
    {
        this.point++;
        datas[point] = data;
    }
    public mynode pop()
    {
        mynode result = datas[this.point];
        this.point--;
        return result;
    }

    public int IsEmpty()
    {
        return point;
    }
}

class mynode
{
    public int id;
    public int find = -1;
    public int finish = -1;
    //隣接するノードを登録する用の変数
    public mynode[] adjacency = null;
    int point = 0;

    public void initialize(int id, int depth)
    {
        this.id = id;
        this.adjacency = new mynode[depth];
    }

    public void RegistAdjacenxy(mynode m)
    {
        this.adjacency[this.point] = m;
        this.point++;
    }
}