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.Collections.Generic;
class Program
    {
        class Node
        {
            int id;
            public int dist;
            public List<Node> links;
            public Node(int i)
            {
                this.id = i;
                this.links = new List<Node>();
                this.dist = -1;
            }
            public void connectTo(Node n)
            {
                this.links.Add(n);
            }
        }
    	static void Main(string[] args)
    	{
            string[] s1 = Console.ReadLine().Split(' ');
            int n = int.Parse(s1[0]), m = int.Parse(s1[1]);
            Node[,] nodes = new Node[n, 3];
            for (int i=0; i<n; i++)
            {
                for (int j=0; j<3; j++)
                {
                    nodes[i, j] = new Node(i + j * 1000000);
                }
            }
            for (int i=0; i<m; i++)
            {
                string[] s2 = Console.ReadLine().Split(' ');
                int u = int.Parse(s2[0]) - 1, v = int.Parse(s2[1]) - 1;
                for (int j=0; j<3; j++)
                {
                    nodes[u, j].connectTo(nodes[v, (j + 1) % 3]);
                }
            }
            Queue<Node> que = new Queue<Node>();
            string[] s3 = Console.ReadLine().Split(' ');
            int s = int.Parse(s3[0]) - 1, t = int.Parse(s3[1]) - 1;
            nodes[s, 0].dist = 0;
            que.Enqueue(nodes[s, 0]);

            while(que.Count > 0)
            {
                Node looking = que.Dequeue();
                foreach (Node nn in looking.links)
                {
                    if (nn.dist < 0)
                    {
                        nn.dist = looking.dist + 1;
                        que.Enqueue(nn);
                    }
                }
            }
            if (nodes[t, 0].dist < 0) Console.WriteLine(-1);
            else Console.WriteLine(nodes[t, 0].dist / 3);
            
        }
    }