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: Wrong Answer)

//use itertools::Itertools;
use std::cmp;
use std::collections::BTreeMap;
use std::collections::BTreeSet;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::VecDeque;
use std::io::Read;
use std::usize::MAX;

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)*}};}
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")};}

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

    let mut adj_list = vec![vec![]; n * 3];
    for &e in &edges {
        let from_base = e.0 * 3;
        let to_base = e.1 * 3;
        adj_list[from_base].push(to_base + 1);
        adj_list[from_base + 1].push(to_base + 2);
        adj_list[from_base + 2].push(to_base);
    }

    let mut visited = vec![false; n * 3];
    let mut walk_count = 0;

    let mut heap = BinaryHeap::new();

    heap.push((s * 3, 0));
    visited[s * 3] = true;

    while !heap.is_empty() {
        let node = heap.pop().unwrap();
        visited[node.0] = true;

        for &i in &adj_list[node.0] {
            if i == t * 3 {
                walk_count = node.1 + 1;
            }
            if !visited[i] {
                heap.push((i, node.1 + 1));
            }
        }

        if walk_count > 0 {
            break;
        }
    }

    if walk_count > 0 {
        println!("{}", walk_count / 3);
    } else {
        println!("{}", -1);
    }
}

fn main() {
    let thd = std::thread::Builder::new().stack_size(104_857_600);
    thd.spawn(|| solve()).unwrap().join().unwrap();

    /*
    // 入力を一括で読み込む場合
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut input = buf.split_whitespace();

    // inputに対しnext()で読み込んでいく
    let q: usize = input.next().unwrap().parse().unwrap();
    */
}

const LARGE_PRIME: u64 = 1_000_000_007;

// @〜でsnippet