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 m = sc.nextInt();

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

        boolean[] v = new boolean[n * 3];
        int[] d = new int[n * 3];
        Arrays.fill(d, Integer.MAX_VALUE);

        Queue<Integer> q = new LinkedList<>();
        q.add(s * 3);
        d[s * 3] = 0;
        v[s*3] = true;
        while(!q.isEmpty()) {
            int a = q.poll();
            int node = a / 3;
            int rem = a % 3;
            int nr = (rem + 1) % 3;
            for(int next : (List<Integer>)e[node]) {
                int na = next * 3 + nr;
                if(v[na]) continue;
                v[na] = true;
                d[na] = Math.min(d[na], d[a] + 1);
                q.add(na);
                if(na == t * 3) {
                    System.out.println(d[na] / 3);
                    return;
                }
            }
        }
        System.out.println(-1);
    }

}
