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

public class Main
{
    public static void main(String[] args)
    {
        Scanner s = new Scanner(System.in);
        int n,m;

        n = s.nextInt();
        m = s.nextInt();

        ArrayList<Integer>[] adj = new ArrayList[3*n+5];
        for(int i=0;i<3*n+5;i++)
            adj[i] = new ArrayList<>();

        int a,b,src,dest;

        for(int i =0;i<m;i++)
        {
            a = s.nextInt();
            b = s.nextInt();

            adj[a].add(b+n);
            adj[a+n].add(b+2*n);
            adj[a+2*n].add(b);
        }

        src = s.nextInt();
        dest = s.nextInt();
        int[] dist = new int[3*n+5];
        for(int i=0;i<3*n+5;i++)
            dist[i]=-1;

        bfs(adj,dist,src,dest,n);

        if(dist[dest]==-1)
            System.out.println(-1);
        else
            System.out.println(dist[dest]/3);

    }

    static void bfs(ArrayList<Integer>[] adj,int[] dist,int src,int dest,int n)
    {
        Queue<Integer> q = new LinkedList<Integer>();
        q.add(src);

        dist[src] =0;
        while(q.size()>0)
        {
            int v = q.peek();
            q.remove();

            for(int u:adj[v])
            {
                if(dist[u]==-1)
                {
                    dist[u]=dist[v]+1;
                    q.add(u);
                }
            }
        }
    }
}