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: Time Limit Exceeded)

import java.util.*;

public class Main{
    static MyPoint mps[];

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        mps = new MyPoint[N+1];
        for(int t=0; t<=N; t++) {
            mps[t] = new MyPoint(t);
        }
        for(int s=1; s<=M; s++) {
            int u = sc.nextInt();
            int v = sc.nextInt();
            mps[u].addEdge(v);
//            mps[v].addEdge(u);
        }
        int S = sc.nextInt();
        int T = sc.nextInt();

        mps[S].check(0, 0);
//        dump(N);
        int tmp = mps[T].minStemps[0];
        if(tmp == Integer.MAX_VALUE) {
            System.out.println(-1);
        } else {
            System.out.println(tmp / 3);
        }
    }

    static void dump(int N) {
        for(int t=1; t<=N; t++) {
            System.out.println(String.format("%d:%d, %d, %d", t, mps[t].minStemps[0], mps[t].minStemps[1], mps[t].minStemps[2]));
        }
    }

    static class MyPoint {
        final int index;
        final Set<Integer> edges = new HashSet<>();
        int minStemps[] = new int[3];

        MyPoint(int i) {
            this.index = i;
            for(int t=0; t<3; t++) {
                minStemps[t] = Integer.MAX_VALUE;
            }
        }
        void addEdge(int x) {
            edges.add(x);
        }

        void check(int mod, int step) {
            mod = mod % 3;
            if(minStemps[mod] <= step) return;
            minStemps[mod] = step;
            int newmod = (mod + 1) % 3;
            for(Integer k : edges) {
                mps[k].check(newmod, step+1);
            }
       }
    }

}