Problem:
Score : 500 points

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

Solution in Rust (1.15.1) (judge verdict: Accepted)

fn main() {
    let (n,m) = {
        let s = rv::<usize>();
        (s[0],s[1])
    };
    let mut g = vec![vec![];n];
    for _ in 0..m {
        let s = rv::<usize>();
        g[s[0]-1].push(s[1]-1);
    }
    let (s,t) = {
        let s = rv::<usize>();
        (s[0]-1,s[1]-1)
    };
    let mut d = vec![[INF;3];n];
    let mut q = std::collections::BinaryHeap::new();
    q.push((0,s));
    while let Some((c,v)) = q.pop() {
        let c = -c;
        let c1 = c/3;
        let c2 = (c%3)as usize;
        if d[v][c2] <= c1 {
            continue;
        }
        d[v][c2]=c1;
        if v==t && c2==0 {
            break;
        }
        let c = c+1;
        let c2 = (c%3)as usize;
        for &u in &g[v] {
            if d[u][c2]>c {
                q.push((-c,u));
            }
        }
    }
    println!("{}",if d[t][0]<INF {d[t][0]} else {-1});
}
const INF: i32 = std::i32::MAX;
fn rl() -> String {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s
}
fn rv<T>() -> Vec<T>
  where T: std::str::FromStr, T::Err: std::fmt::Debug, {
    rl().split_whitespace().map(|x|x.parse::<T>().unwrap()).collect()
}