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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Text;
using System.Collections.Generic;
using static System.Math;
using System.Linq;
using static System.Array;

namespace ConsoleApp2
{
    class Program
    {
        static long N;
        static long M;
        static long S;
        static long T;
        static Node[] Nodes;
        static void Main(string[] args)
        {
            var NM = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();
            N = NM[0];
            M = NM[1];
            Nodes = new Node[3 * N + 1];

            for (int i = 0; i < 3 * N + 1; i++)
            {
                Nodes[i] = new Node();
            }

            for (int i = 0; i < M; i++)
            {
                var UV = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();
                var U = UV[0];
                var V = UV[1];

                Nodes[3 * U].adj.Add(3 * V - 1);
                Nodes[3 * U - 1].adj.Add(3 * V - 2);
                Nodes[3 * U - 2].adj.Add(3 * V);
            }
            var ST = Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToArray();
            S = ST[0];
            T = ST[1];

            Console.WriteLine(WFS());
        }

        static long WFS()
        {
            var que = new Queue<long>();
            que.Enqueue(3 * S);
            Nodes[3 * S].dis = 0;
            while(que.Count > 0)
            {
                var top = que.Dequeue();
                var now = Nodes[top];
                foreach (var item in now.adj)
                {
                    if(Nodes[item].dis == -1)
                    {
                        Nodes[item].dis = now.dis + 1;
                        que.Enqueue(item);
                    }
                }
            }
            return Nodes[3*T].dis%3 == 0 ? Nodes[3*T].dis/3 : -1;
        }
    }

    class Node
    {
        public List<long> adj;
        public long dis = -1;

        public Node()
        {
            adj = new List<long>();
        }
    }
}