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: Time Limit Exceeded)

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        #[cfg(debug_assertions)]
        writeln!(&mut std::io::stderr(), concat!("[DEBUG] ", $(stringify!($a), "={:?} "),*), $($a),*);
    }
}

#[allow(unused_imports)]
use std::cmp::{min, max};

#[allow(unused_imports)]
use std::io::Write;
use std::cmp::Ordering;

#[derive(Debug, Copy, Clone, Eq, PartialEq)]
struct Rev(usize);

impl Ord for Rev {
    fn cmp(&self, other: &Rev) -> Ordering {
        other.0.cmp(&self.0)
    }
}

impl PartialOrd for Rev {
    fn partial_cmp(&self, other: &Rev) -> Option<Ordering> {
        Some(other.0.cmp(&self.0))
    }
}


fn kenken(edges: &Vec<Vec<usize>>, root: usize, cur: usize, d: usize, e2: &mut Vec<Vec<usize>>) {
    if d == 3 {
        e2[root].push(cur);
        return;
    }

    for &nd in edges[cur].iter() {
        kenken(edges, root, nd, d+1, e2);
    }
}


fn main() {
    input!{
      n: usize,
      m: usize,
      uvs: [(usize1,usize1); m],
      s: usize1,
      t: usize1
    }

    let mut edges = vec![vec![]; n];
    for (u, v) in uvs {
        edges[u].push(v);
    }

    let mut edges2 = vec![vec![]; n];
    for i in 0..n {
        kenken(&edges, i, i, 0, &mut edges2);
    }
    let edges = edges2;
    use std::collections::BinaryHeap;

    // ダイクストラ
    const INF: usize = 1 << 60;
    let mut dist = vec![INF; n];
    let mut q = BinaryHeap::new();
    q.push((Rev(0),s));
    dist[s] = 0;

    while q.len() > 0 {
        let (Rev(d), cur) = q.pop().unwrap();
        if dist[cur] < d {
            continue;
        }

        for &nd in edges[cur].iter() {
            if dist[nd] > dist[cur] + 1 {
                dist[nd] = dist[cur] + 1;
                q.push((Rev(dist[nd]), nd));
            }
        }
    }

    let ans = dist[t];
    if ans >= INF {
        println!("{}", -1);
    } else {
        println!("{}", ans);
    }

}