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) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long m = sc.nextLong();
        
        // to: 隣接リスト(有向)
        List<Integer>[] to = new List[n];
        for (int i = 0; i < n; i++) {
            to[i] = new ArrayList<Integer>();
        }
        for (int i = 0; i < m; i++) {
            int p = sc.nextInt() - 1;
            int q = sc.nextInt() - 1;
            to[p].add(q);
        }
        
        int start = sc.nextInt() - 1;
        int goal = sc.nextInt() - 1;
        
        // ans: ビルstartの0階からビルgoalの0階に行くまでの歩数
        // 但し、ビル移動毎に1階上がる(二次元の最短経路問題)
        int ans = 0;
        Queue<int[]> que = new ArrayDeque<int[]>();
        que.add(new int[]{start, 0, 0});
        boolean[][] visited = new boolean[n][3];
        while (!que.isEmpty()) {
            int[] cur = que.poll();
            int p = cur[0];
            int f = cur[1];
            int l = cur[2];
            
            if (visited[p][f] == true) continue;
            visited[p][f] = true;
            
            if (p == goal && f == 0) {
                ans = l;
                break;
            }
            
            // 子の次の階に歩数+1を配る
            for (Integer q : to[p]) {
                que.add(new int[]{q, (f + 1) % 3, l + 1});
            }
        }
        if (ans == 0) {
            System.out.println(-1);
        } else {
            // 3歩 = けんけんぱ1回分
            System.out.println(ans / 3);
        }
    }
}