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.ArrayDeque;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int m = in.nextInt();
		Map<Integer,List<Integer>> map = new HashMap<>();
		for(int i=0;i<m;i++) {
			int u = in.nextInt();
			int v = in.nextInt();
			if(map.containsKey(u)) map.get(u).add(v);
			else {
				List<Integer> list = new ArrayList<>();
				list.add(v);
				map.put(u, list);
			}
		}
		int s = in.nextInt();
		int t = in.nextInt();
		int[][] dist = new int[n+1][3];
		for(int i=0;i<n+1;i++) {
			for(int j=0;j<3;j++) {
				dist[i][j] = 100000000;
			}
		}
		dist[s][0] = 0;
		boolean[][] used = new boolean[n+1][3];
		used[s][0] = true;
		Queue<int[]> deq = new ArrayDeque<>();
		int[] start = {s,0};
		deq.add(start);
		while(!deq.isEmpty()) {
			int[] tmp = deq.poll();
			if(!map.containsKey(tmp[0])) continue;
			for(int next : map.get(tmp[0])) {
				dist[next][(tmp[1]+1)%3] = Math.min(dist[tmp[0]][tmp[1]]+1, dist[next][(tmp[1]+1)%3]);
				if(!used[next][(tmp[1]+1)%3]) {
					int[] tmp2 = {next,(tmp[1]+1)%3};
					deq.add(tmp2);
					used[next][(tmp[1]+1)%3] = true;
				}
			}
		}
		if(dist[t][0]==100000000) System.out.println(-1);
		else System.out.println(dist[t][0]/3);
		
		in.close();

	}

}