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<Integer>[] g = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            g[i] = new ArrayList<>();
        }
        for (int i = 0; i < m; i++) {
            int v = sc.nextInt() - 1;
            int u = sc.nextInt() - 1;
            g[v].add(u);
        }
        int s = sc.nextInt() - 1;
        int t = sc.nextInt() - 1;
        int[][] d = new int[n][3];
        for (int i = 0; i < d.length; i++) {
            Arrays.fill(d[i], Integer.MAX_VALUE);
        }

        Queue<int[]> q = new LinkedList<>();
        q.offer(new int[]{s, 0});
        d[s][0] = 0;

        while (!q.isEmpty()) {
            int[] vx = q.poll();
            int v = vx[0];
            int state = vx[1];
            int next = (vx[1] + 1) % 3;
            for (int i = 0; i < g[v].size(); i++) {
                if (d[g[v].get(i)][next] != Integer.MAX_VALUE) continue;
                d[g[v].get(i)][next] = d[v][state] + 1;
                q.offer(new int[]{g[v].get(i), next});
            }
        }

        if (d[t][0] == Integer.MAX_VALUE) {
            System.out.println(-1);
            return;
        }

        System.out.println(d[t][0] / 3);
    }
}