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.*;

public class Main {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int mm = sc.nextInt();

        int[] u = new int[mm];
        int[] v = new int[mm];
        for(int i = 0; i < mm; i++) {
            u[i] = sc.nextInt();
            v[i] = sc.nextInt();
        }
        int s = sc.nextInt();
        int t = sc.nextInt();
        ArrayList<ArrayList<Integer>> l = new ArrayList<>();
        for(int i = 0; i < n; i++) {
            l.add(new ArrayList<>());
        }
        for(int i = 0; i < mm; i++) {
            l.get(u[i] - 1).add(v[i] - 1);
        }

        boolean[][] visit = new boolean[n][4];
        ArrayDeque<Move> q = new ArrayDeque<>();
        q.add(new Move(s-1, 0, 0));
        int ans = -1;
        while(!q.isEmpty()) {
            Move m = q.poll();
            if(visit[m.now][m.num]) {
                continue;
            }
            visit[m.now][m.num] = true;
            if(m.num == 0) {
                if(m.now == t-1) {
                    ans = m.cc;
                    break;
                }
                m.num = 3;
                m.cc++;
            } 
            ArrayList<Integer> ll = l.get(m.now);

            for(int i = 0; i < ll.size(); i++) {
                q.add(new Move(ll.get(i), m.num - 1, m.cc));
            }
        }

        System.out.println(ans);



    }
    static class Move {
        int now;
        int num;
        int cc;
        Move(int a, int b, int c) {
            now = a;
            num = b;
            cc = c;
        }
    }
}