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();
        int[][] g = buildGraph(sc, n, m);
        int s = sc.nextInt()-1;
        int t = sc.nextInt()-1;
        int[][] a = new int[n][3];
        for (int i = 0; i < n; i++) {
            Arrays.fill(a[i], -1);
        }
        ArrayDeque<int[]> ad = new ArrayDeque<>();
        ad.add(new int[]{s, 0});
        while(!ad.isEmpty()){
            int v = ad.peek()[0];
            int l = ad.peek()[1];
            int nl =( l + 1) % 3;
            ad.poll();
            for(int next : g[v]){
                if(a[next][nl] != -1) continue;
                a[next][nl] = l + 1;
                ad.add(new int[]{next, l+1});
            }
        }
        System.out.println(a[t][0] == -1 ? -1 : a[t][0] / 3);
        sc.close();

    }

    public static int[][] buildGraph(Scanner sc, int n, int m){
        int[][] graph = new int[n][];
        int[][] edges = new int[m][2];
        int[] deg = new int[n];
        for (int i = 0; i < m; i++) {
            int a = sc.nextInt() -1;
            int b = sc.nextInt() -1;
            edges[i][0] = a;
            edges[i][1] = b;
            deg[a]++;
        }
        for (int i = 0; i < n; i++) {
            graph[i] = new int[deg[i]];
        }
        for (int i = 0; i < m; i++) {
            int a = edges[i][0];
            int b = edges[i][1];
            graph[a][--deg[a]] = b;
        }

        return graph;
    }

}

