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.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Deque;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

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

        final Map<Integer, List<Integer>> graph = new HashMap<>();
        for (int i = 0; i < m; i++) {
            final int u = scanner.nextInt();
            final int v = scanner.nextInt();
            graph.computeIfAbsent(u * 3, list -> new ArrayList<>()).add(v * 3 + 1);
            graph.computeIfAbsent(u * 3 + 1, list -> new ArrayList<>()).add(v * 3 + 2);
            graph.computeIfAbsent(u * 3 + 2, list -> new ArrayList<>()).add(v * 3);
        }

        final int s = scanner.nextInt() * 3;
        final int t = scanner.nextInt() * 3;
        final int[] distances = new int[(n + 1) * 3];
        Arrays.fill(distances, -1);
        distances[s] = 0;
        final Deque<Integer> queue = new ArrayDeque<>();
        queue.add(s);
        while (!queue.isEmpty()) {
            final int current = queue.pollFirst();
            final List<Integer> list = graph.get(current);
            if (list == null) {
                continue;
            }
            for (final int next : list) {
                if (distances[next] >= 0) {
                    continue;
                }
                distances[next] = distances[current] + 1;
                queue.add(next);
            }
        }

        System.out.println(distances[t] >= 0 ? distances[t] / 3 : -1);
    }
}
