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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.Arrays;
import java.util.AbstractCollection;
import java.util.Map;
import java.util.Scanner;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Collections;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        E solver = new E();
        solver.solve(1, in, out);
        out.close();
    }

    static class E {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int N = in.nextInt();
            int M = in.nextInt();
            Map<Integer, List<Integer>> map = new HashMap<>();
            for (int i = 0; i < M; i++) {
                int u = in.nextInt();
                map.put(u, map.getOrDefault(u, new ArrayList<>()));
                map.get(u).add(in.nextInt());
            }
            int S = in.nextInt();
            int T = in.nextInt();

            int[][] memo;
            memo = new int[N + 1][4];
            for (int i = 0; i < N + 1; i++) {
                Arrays.fill(memo[i], -1);
            }
            memo[S][0] = 0;

            LinkedList<List<Integer>> que = new LinkedList<>();
            que.add(Arrays.asList(S, 0));
            while (!que.isEmpty()) {
                List<Integer> l = que.removeFirst();
                int s = l.get(0);
                int kenkenpa = l.get(1);
                if (s == T && kenkenpa == 3) {
                    out.println(memo[s][kenkenpa] / 3);
                    return;
                }
                int temp = (kenkenpa % 3) + 1;
                for (int u : map.getOrDefault(s, Collections.emptyList())) {
                    if (memo[u][temp] > 0) {
                        continue;
                    }
                    memo[u][temp] = memo[s][kenkenpa] + 1;
                    que.add(Arrays.asList(u, temp));
                }
            }
            out.println(-1);
        }

    }
}

