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)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
use std::io::{Write, BufWriter};
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    ($($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)*}
    };
}

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)*}
    };
}

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

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

/*
 * Dijkstra's algorithm.
 * Verified by: AtCoder ABC035 (http://abc035.contest.atcoder.jp/submissions/676539)
 */

struct Dijkstra {
    edges: Vec<Vec<(usize, i64)>>, // adjacent list representation
}

/*
 * Code from https://doc.rust-lang.org/std/collections/binary_heap/
 */
#[derive(Copy, Clone, Eq, PartialEq)]
struct State {
    cost: i64,
    position: usize,
}

// The priority queue depends on `Ord`.
// Explicitly implement the trait so the queue becomes a min-heap
// instead of a max-heap.
impl Ord for State {
    fn cmp(&self, other: &State) -> Ordering {
        // Notice that the we flip the ordering here
        match other.cost.cmp(&self.cost) {
            std::cmp::Ordering::Equal => other.position.cmp(&self.position),
            x => x,
        }
    }
}

// `PartialOrd` needs to be implemented as well.
impl PartialOrd for State {
    fn partial_cmp(&self, other: &State) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}


impl Dijkstra {
    fn new(n: usize) -> Self {
        Dijkstra { edges: vec![Vec::new(); n] }
    }
    fn add_edge(&mut self, from: usize, to: usize, cost: i64) {
        self.edges[from].push((to, cost));
    }
    /*
     * This function returns a Vec consisting of the distances from vertex source.
     */
    fn solve(&self, source: usize, inf: i64) -> Vec<i64> {
        let n = self.edges.len();
        let mut d = vec![inf; n];
        let mut que = std::collections::BinaryHeap::new();
        que.push(State {cost: 0, position: source});
        while let Some(State {cost, position: pos}) = que.pop() {
            if d[pos] <= cost {
                continue;
            }
            d[pos] = cost;
            for adj in &self.edges[pos] {
                que.push(State {cost: cost + adj.1, position: adj.0});
            }
        }
        return d;
    }
}

fn solve() {
    let out = std::io::stdout();
    let mut out = BufWriter::new(out.lock());
    macro_rules! puts {
        ($($format:tt)*) => (write!(out,$($format)*).unwrap());
    }
    input! {
        n: usize, m: usize,
        uv: [(usize1, usize1); m],
        s: usize1, t: usize1,
    }
    let mut dijk = Dijkstra::new(3 * n);
    for &(u, v) in &uv {
        for d in 0..3 {
            let e = (d + 1) % 3;
            dijk.add_edge(u + d * n, v + e * n, 1);
        }
    }
    const INF: i64 = 1 << 50;
    let ans = dijk.solve(s, INF)[t];
    puts!("{}\n", if ans >= INF { -1 } else { ans / 3 });
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}