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) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		List<Integer>[] edges = new List[N];
		for (int i = 0; i < N; i++) {
			edges[i] = new LinkedList<>();
		}
		for (int i = 0; i < M; i++) {
			edges[sc.nextInt() - 1].add(sc.nextInt() - 1);
		}
		int s = sc.nextInt() - 1;
		int t = sc.nextInt() - 1;
		int[][] dp = new int[N][3];
		for (int i = 0; i < N; i++) {
			Arrays.fill(dp[i], 1 << 25);
		}
		dp[s][0] = 0;
		Queue<int[]> Q = new LinkedList<>();
		Q.add(new int[] {s, 0});
		while (!Q.isEmpty()) {
			int u = Q.peek()[0];
			int d = Q.poll()[1];
			if (u == t && d == 0) {
				System.out.println(dp[u][d]/3);
				return;
			}
			for (int v: edges[u]) {
				if (dp[v][(d + 1) % 3] > dp[u][d] + 1) {
					dp[v][(d + 1) % 3] = dp[u][d] + 1;
					Q.add(new int[] {v, (d + 1) % 3});
				}
			}
		}
		System.out.println(-1);
	}
}