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.*;
import java.util.stream.*;
import java.io.*;

public class Main {

  public static void main(String[] args) throws IOException {

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

    String[] in1 = br.readLine().split(" ");
    int N = Integer.parseInt(in1[0]);
    int M = Integer.parseInt(in1[1]);

    List<Integer>[] graph = new List[3 * N];
    for(int i = 0; i < 3 * N; i++) graph[i] = new ArrayList<Integer>();

    for(int i = 0; i < M; i++){
      in1 = br.readLine().split(" ");
      int u = Integer.parseInt(in1[0]);
      int v = Integer.parseInt(in1[1]);
      u--; v--;
      graph[3 * u].add(3 * v + 1);
      graph[3 * u + 1].add(3 * v + 2);
      graph[3 * u + 2].add(3 * v);
    }

    in1 = br.readLine().split(" ");
    int S = Integer.parseInt(in1[0]);
    int T = Integer.parseInt(in1[1]);
    S--;
    T--;
    S *= 3;
    T *= 3;

    br.close();

    int[] minPath = new int[3 * N];
    Arrays.fill(minPath, -1);
    minPath[S] = 0;

    Queue<Integer> que = new ArrayDeque<>();
    que.add(S);

    while(!que.isEmpty()){

      int edge = que.poll();

      for(int next : graph[edge]){
        if(minPath[next] != -1) continue;

        minPath[next] = minPath[edge] + 1;
        que.add(next);
      }
    }

    int ans = minPath[T];
    if(ans == -1 || ans % 3 != 0) System.out.println(-1);
    else System.out.println(ans / 3);
  }
}