Problem Statement

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 $T$.

Sample Input 1

4 4
1 2
2 3
3 4
4 1
1 3

Sample Output 1

2

Ken can reach Vertex $3$ from Vertex $1$ in two ken-ken-pa, as follows: $1 \\rightarrow 2 \\rightarrow 3 \\rightarrow 4$ in the first ken-ken-pa, then $4 \\rightarrow 1 \\rightarrow 2 \\rightarrow 3$ in the second ken-ken-pa. This is the minimum number of ken-ken-pa needed.

Sample Input 2

3 3
1 2
2 3
3 1
1 2

Sample Output 2

-1

Any number of ken-ken-pa will bring Ken back to Vertex $1$, so he cannot reach Vertex $2$, though he can pass through it in the middle of a ken-ken-pa.

Sample Input 3

2 0
1 2

Sample Output 3

-1

Vertex $S$ and Vertex $T$ may be disconnected.

Sample Input 4

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6

Sample Output 4

2