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: Wrong Answer)

using System;
using System.Collections.Generic;
class Program
{
	static void Main(string[] args)
	{
		string[] str = Console.ReadLine().Split();
		int N = int.Parse(str[0]);
		int M = int.Parse(str[1]);
		int[][]A = new int[M][];
		var listA = new HashSet<int>[N];
		for(var i=0;i<N;i++){
			listA[i] = new HashSet<int>();
		}
		for(var i=0;i<M;i++){
			string[] str2 = Console.ReadLine().Split();
			A[i] = new int[2]{int.Parse(str2[0]),int.Parse(str2[1])};
			listA[A[i][0]-1].Add(A[i][1]);
		}
		string[] str3 = Console.ReadLine().Split();
		int S = int.Parse(str3[0]);
		int T = int.Parse(str3[1]);
		var list = new HashSet<int>();
		list.Add(S);
		int[]C = new int[N];
		C[S-1] = 1;
		int ans = 0;
		var listB = new HashSet<int>[N];
		for(var i=0;i<N;i++){
			listB[i] = new HashSet<int>();
		}
		while(list.Count>0){
				ans++;
				int[]D = new int[N];
				var list2 = new HashSet<int>(list);
				list.Clear();
				bool flag = false;
				bool flag2 = false;
				foreach(var i in list2){
					if(listB[i-1].Count==0){
					foreach(var j in listA[i-1]){
						list.Add(j);
					}
					var list3 = new HashSet<int>(list);
					list.Clear();
					foreach(var i2 in list3){
					foreach(var j in listA[i2-1]){
						list.Add(j);
					}
					}
					list3 = new HashSet<int>(list);
					list.Clear();
					foreach(var i2 in list3){
					foreach(var j in listA[i2-1]){
						if(C[j-1]==0){
							list.Add(j);
							listB[i-1].Add(j);
							C[j-1]=1;
							flag = true;
							if(j==T){
								flag2 = true;
								break;
							}
						}
					}
					}
					} else {
					foreach(var j in listB[i-1]){
						if(C[j-1]==0){
							C[j-1]=1;
							flag = true;
							if(j==T){
								flag2 = true;
								break;
							}
						}
					}
					}
				}
				if(flag2){
					break;
				}
				if(!flag){
					ans = -1;
					break;
				}
		}
		Console.WriteLine(ans);
	}
}