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 System.Linq;
using System.Numerics;

class AtCoder
{
    const int MOD = 1000000007;
    
    public void Slove()
    {
        var inputs = Console.ReadLine().Split();

        var n = int.Parse(inputs[0]);
        var m = int.Parse(inputs[1]);
        var edges = 
            Enumerable.Range(0, n)
            .Select(x => new List<int>())
            .ToArray();
        foreach(int i in Enumerable.Range(0, m))
        {
            inputs = Console.ReadLine().Split();
            var u = int.Parse(inputs[0]) - 1;
            var v = int.Parse(inputs[1]) - 1;
            edges[u].Add(v);
        }
        inputs = Console.ReadLine().Split();
        var s = int.Parse(inputs[0]) - 1;
        var t = int.Parse(inputs[1]) - 1;
        var dp = 
            Enumerable.Range(0, 3)
            .Select(x => new bool[n])
            .ToArray(); 
        dp[0][s] = true;
        var que = new Queue<int>();
        que.Enqueue(s);
        int cnt = 0;
        while(que.Any() && !dp[0][t])
        {
            int nmod = (++cnt) % 3;
            int queCount = que.Count;
            foreach(int i in Enumerable.Range(0, queCount))
            {
                int current = que.Dequeue();
                foreach(var next in edges[current].Where(x => !dp[nmod][x]))
                {
                    dp[nmod][next] = true;
                    //System.Console.WriteLine($"{next} {nmod}");
                    que.Enqueue(next);
                }
            }
        }
        if(dp[0][t])
            System.Console.WriteLine(cnt / 3);
        else
        {
            System.Console.WriteLine(-1);
        }
        // System.Console.WriteLine(string.Join(" ", dp[0]));
        // System.Console.WriteLine(string.Join(" ", edges[0]));
        //     System.Console.WriteLine(t);
    }
    static void Main()
    {
        var ac = new AtCoder();
        ac.Slove();
    }
}