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.

Input is given from Standard Input in the following format:

```
$N$ $M$
$u_1$ $v_1$
$:$
$u_M$ $v_M$
$S$ $T$
```

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 $T$.

-   $2 \leq N \leq 10^5$
-   $0 \leq M \leq \min(10^5, N (N-1))$
-   $1 \leq u_i, v_i \leq N(1 \leq i \leq M)$
-   $u_i \neq v_i (1 \leq i \leq M)$
-   If $i \neq j$, $(u_i, v_i) \neq (u_j, v_j)$.
-   $1 \leq S, T \leq N$
-   $S \neq T$

2

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

-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.

-1

Vertex S and Vertex T may be disconnected.