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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
public class Main {

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int m = Integer.parseInt(st.nextToken());
		ArrayList<Integer>[] g = new ArrayList[n];
		for(int i = 0; i < n; i++) g[i] = new ArrayList<>();
		for(int i = 0; i < m; i++){
			st = new StringTokenizer(br.readLine());
			int u = Integer.parseInt(st.nextToken())-1;
			int v = Integer.parseInt(st.nextToken())-1;
			g[u].add(v);
		}
		st = new StringTokenizer(br.readLine());
		int s = Integer.parseInt(st.nextToken())-1;
		int t = Integer.parseInt(st.nextToken())-1;
		int[][] dist = new int[3][n];
		for(int i = 0; i < 3; i++) Arrays.fill(dist[i], -1);
		dist[0][s] = 0;
		LinkedList<Integer> q = new LinkedList<>();
		q.add(s);
		while(!q.isEmpty()){
			int u = q.removeFirst();
			for(int v: g[u]){
				boolean yes = false;
				for(int d = 0; d < 3; d++){
					if(dist[d][u] >= 0 && dist[(d+1)%3][v] < 0){
						dist[(d+1)%3][v] = dist[d][u] + 1;
						yes = true;
					}
				}
				if(yes) q.add(v);
			}
		}
		System.out.println(dist[0][t] >= 0 ? dist[0][t] / 3 : dist[0][t]);
	}
}