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: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ABC132E3
{
    class Program
    {
        static long[] dp;
        static long[,] map;
        static int count = 0;
        static void Main(string[] args)
        {
            string[] input = Console.ReadLine().Split();
            long N = long.Parse(input[0]);
            long M = long.Parse(input[1]);
            if (M == 0)
            {
                Console.WriteLine("-1");
            }
            else
            {
                long[] u = new long[M + 1];
                long[] v = new long[M + 1];
                map = new long[M + 1, M + 1];
                for (int i = 1; i <= M; i++)
                {
                    string[] input2 = Console.ReadLine().Split();
                    u[i] = long.Parse(input2[0]);
                    v[i] = long.Parse(input2[1]);
                    map[u[i], v[i]] = 1;
                }
                string[] input3 = Console.ReadLine().Split();
                int S = int.Parse(input3[0]);
                int T = int.Parse(input3[1]);
                dp = new long[N + 1];
                for (int i = 0; i <= N; i++)
                {
                    dp[i] = 0;
                }
                long ans = trace(N, S, T, T);
                if (ans != -1)
                {
                    ans = ans / 3;
                }
                Console.WriteLine(ans);
            }
        }
        static long trace(long N, long S, int goal, int T)
        {
            for (int i = 1; i <= N; i++)
            {
                if (map[i, goal] == 1)
                {
                    count++;
                    dp[i] = count;
                    if ((dp[T] != 0) && (dp[T] % 3 == 0))
                    {
                        //到達できない時
                        count = -1;
                        return count;
                    }
                    if ((dp[S] != 0) && (dp[S] % 3 == 0))
                    {
                        return count;
                    }
                    else
                    {
                        trace(N, S, i, T);
                    }
                }
            }
            return count;
        }
    }
}