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.awt.*;
import java.util.*;

public class Main {
    static Scanner scanner = new Scanner(System.in);
    static int mod = 1000000007;

    public static void main(String[]$) {
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        Map<Integer, Set<Integer>> edge = new HashMap<>();
        for (int i = 0; i < n; i++) {
            edge.put(i, new HashSet<>());
        }
        for (int i = 0; i < m; i++) {
            edge.get(scanner.nextInt() - 1).add(scanner.nextInt() - 1);
        }
        int[][] can = new int[n][3];
        int s = scanner.nextInt() - 1;
        int t = scanner.nextInt() - 1;
        can[s][0] = 1;
        Deque<Point> deque = new ArrayDeque<>();
        deque.add(new Point(s, 0));
        while (!deque.isEmpty()) {
            Point point = deque.poll();
            for (int v : edge.get(point.x)) {
                if (can[v][(point.y + 1) % 3] == 0) {
                    can[v][(point.y + 1) % 3] = can[point.x][point.y] + 1;
                    deque.add(new Point(v, (point.y + 1) % 3));
                } else {
                    can[v][(point.y + 1) % 3] = Math.min(can[v][(point.y + 1) % 3], can[point.x][point.y] + 1);
                }
            }
        }
        if (can[t][0] == 0) {
            System.out.println(-1);
            return;
        }
        System.out.println(can[t][0] / 3);
    }
}