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.

Examples

Input

4 4
1 2
2 3
3 4
4 1
1 3


Output

2


Input

3 3
1 2
2 3
3 1
1 2


Output

-1


Input

2 0
1 2


Output

-1


Input

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6


Output

2
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Queue;

public class Main {

	public static void main(String args[]) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String str[] = br.readLine().split(" ");

		int n = Integer.parseInt(str[0]);
		int m = Integer.parseInt(str[1]);

		ArrayList<ArrayList<Integer>> edge = new ArrayList<>();

		for (int i = 0; i < n; i++) {
			edge.add(new ArrayList<>());
		}

		for (int i = 0; i < m; i++) {
			str = br.readLine().split(" ");
			int u = Integer.parseInt(str[0]) - 1;
			int v = Integer.parseInt(str[1]) - 1;

			edge.get(u).add(v);
		}

		str = br.readLine().split(" ");
		int s = Integer.parseInt(str[0]) - 1;
		int t = Integer.parseInt(str[1]) - 1;

		int dist[][] = new int[n][3];
		Queue<Integer> queueV = new ArrayDeque<>();
		Queue<Integer> queueK = new ArrayDeque<>();

		for (int i = 0; i < n; i++) {
			Arrays.fill(dist[i], Integer.MAX_VALUE);
		}

		dist[s][0] = 0;
		queueV.add(s);
		queueK.add(0);

		while (!queueV.isEmpty() || !queueK.isEmpty()) {
			int v = queueV.poll();
			int k = queueK.poll();

			for (int u : edge.get(v)) {
				int nextk = (k + 1) % 3;
				if (dist[u][nextk] != Integer.MAX_VALUE)
					continue;
				dist[u][nextk] = dist[v][k] + 1;
				queueV.add(u);
				queueK.add(nextk);
			}
		}

		int ans = dist[t][0];
		if (ans == Integer.MAX_VALUE)
			ans = -1;
		else
			ans /= 3;
		System.out.println(ans);
	}
}
