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.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Scanner;

public class Main {

    static class Node {

        int num;

        int dist;

        public Node(int num, int dist) {

            this.num = num;

            this.dist = dist;

        }
    }

    static class Vertex {

        int num;

        List<Integer> edges = new LinkedList<>();

        public Vertex(int num) {

            this.num = num;
        }
    }

    static Vertex[] nodes;

    static int N;

    static int M;

    static int S;

    static int T;
    public static void main(String[] args) {

        Scanner s = new Scanner(System.in);

        N = s.nextInt();

        M = s.nextInt();

        nodes = new Vertex[N+1];

        for (int i = 1; i <= N; i++) {

            nodes[i] = new Vertex(i);
        }

        for (int i = 0; i < M; i++) {

            int from = s.nextInt();

            int to = s.nextInt();


            nodes[from].edges.add(to);
        }

        S = s.nextInt();

        T = s.nextInt();

        bfs();

    }

    public static void bfs() {


        boolean[][] flag = new boolean[N+1][3];


        Queue<Node> q = new LinkedList<>();

        flag[S][0] = true;

        Node n = new Node(S, 0);

        q.add(n);

            while (q.isEmpty() == false) {

                Node k = q.poll();

                int dist = k.dist;

            if (dist % 3 ==0 && k.num == T) {

                System.out.println(dist/3);

                return;
            }

            int newDist = dist + 1;


            List<Integer> list = nodes[k.num].edges;

            for (int i : list) {

                if (flag[i][newDist %3] == true) continue;

                flag[i][newDist%3] = true;

                q.add(new Node(i, newDist));

            }

        }

        System.out.println(-1);
    }


}
