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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

// practice with kaiboy
import java.io.*;
import java.util.*;

public class Main extends PrintWriter {
	Main() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		Main o = new Main(); o.main(); o.flush();
	}

	int[] oo; int[][] oj;
	void link(int i, int j) {
		int o = oo[i];
		if (o >= 2 && (o & o - 1) == 0)
			oj[i] = Arrays.copyOf(oj[i], o << 1);
		oj[i][oo[i]++] = j;
	}
	void init(int n, int m) {
		oo = new int[n]; oj = new int[n][2];
	}
	void main() {
		int n = sc.nextInt();
		int m = sc.nextInt();
		init(n, m);
		while (m-- > 0) {
			int i = sc.nextInt() - 1;
			int j = sc.nextInt() - 1;
			link(i, j);
		}
		int s = sc.nextInt() - 1;
		int t = sc.nextInt() - 1;
		int u = s * 3 + 0;
		int v = t * 3 + 0;
		int inf = n * 3;
		int[] qu = new int[n * 3]; int head = 0, cnt = 0;
		int[] dd = new int[n * 3]; Arrays.fill(dd, inf);
		dd[u] = 0; qu[head + cnt++] = u;
		while (cnt > 0) {
			u = qu[head++]; cnt--;
			int d = dd[u] + 1, i = u / 3, r = (u + 1) % 3;
			for (int o = 0; o < oo[i]; o++) {
				int j = oj[i][o];
				u = j * 3 + r;
				if (dd[u] == inf) {
					dd[u] = d;
					if (u == v) {
						println(d / 3);
						return;
					}
					qu[head + cnt++] = u;
				}
			}
		}
		println(-1);
	}
}