Problem:
Score : 500 points

Problem Statement

Ken loves ken-ken-pa (Japanese version of hopscotch). Today, he will play it on a directed graph G.
G consists of N vertices numbered 1 to N, and M edges. The i-th edge points from Vertex u_i to Vertex v_i.

First, Ken stands on Vertex S. He wants to reach Vertex T by repeating ken-ken-pa. In one ken-ken-pa, he does the following exactly three times: follow an edge pointing from the vertex on which he is standing.

Determine if he can reach Vertex T by repeating ken-ken-pa. If the answer is yes, find the minimum number of ken-ken-pa needed to reach Vertex T. Note that visiting Vertex T in the middle of a ken-ken-pa does not count as reaching Vertex T by repeating ken-ken-pa.

Constraints

2 \leq N \leq 10^5

0 \leq M \leq \min(10^5, N (N-1))

1 \leq u_i, v_i \leq N(1 \leq i \leq M)

u_i \neq v_i (1 \leq i \leq M)

If i \neq j, (u_i, v_i) \neq (u_j, v_j).

1 \leq S, T \leq N

S \neq T

Input

Input is given from Standard Input in the following format:

N M
u_1 v_1
:
u_M v_M
S T

Output

If Ken cannot reach Vertex T from Vertex S by repeating ken-ken-pa, print -1.
If he can, print the minimum number of ken-ken-pa needed to reach vertex T.

Sample Input 1

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

Sample Output 1

2

Ken can reach Vertex 3 from Vertex 1 in two ken-ken-pa, as follows: 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 in the first ken-ken-pa, then 4 \rightarrow 1 \rightarrow 2 \rightarrow 3 in the second ken-ken-pa. This is the minimum number of ken-ken-pa needed.

Sample Input 2

3 3
1 2
2 3
3 1
1 2

Sample Output 2

-1

Any number of ken-ken-pa will bring Ken back to Vertex 1, so he cannot reach Vertex 2, though he can pass through it in the middle of a ken-ken-pa.

Sample Input 3

2 0
1 2

Sample Output 3

-1

Vertex S and Vertex T may be disconnected.

Sample Input 4

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

Sample Output 4

2

Solution in C# (.NET Core 3.1.201) (judge verdict: Time Limit Exceeded)

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

namespace atcoder
{
    class Program
    {
        static List<List<int>> graph;
        const int N = 100000;
        static void Main(string[] args)
        {
            var input = Console.ReadLine().Split(" ").Select(s => long.Parse(s)).ToList();
            long n = input[0], m = input[1];

            graph = new List<List<int>>();
            for (var i = 0; i < 3 * N; i++)
            {
                graph.Add(new List<int>());
            }
            for (var i = 0; i < m; i++)
            {
                var line = Console.ReadLine().Split(" ").Select(s => Int32.Parse(s)).ToList();
                var u = line[0] - 1;
                var v = line[1] - 1;
                graph[u].Add(v + N);
                graph[u + N].Add(v + 2 * N);
                graph[u + 2 * N].Add(v);
            }

            var l = Console.ReadLine().Split(" ").Select(s => Int32.Parse(s)).ToList();
            int s = l[0] - 1, t = l[1] - 1;

            DPS(s);
            if (distance[t] == int.MaxValue)Console.WriteLine(-1);
            else Console.WriteLine(distance[t] / 3);
        }

        static List<bool> isChecked = new List<bool>();
        static long[] distance;
        static void DPS(int index)
        {
            isChecked = new List<bool>(new bool[3 * N]);
            distance = new long[3 * N];
            for (int i = 0; i < distance.Length; i++)distance[i] = int.MaxValue;

            _DPS(index, 0);
            return;
        }
        static void _DPS(int index, int d)
        {
            if(distance[index] < d) return;
            foreach (var g in graph[index])
            {
                if (d + 1 < distance[g])
                {
                    distance[g] = d + 1;
                    _DPS(g, d + 1);
                }
            }
        }
    }
}