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: Runtime Error)

#[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>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

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

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

#[allow(dead_code)]
struct UnionFind {
    parent: Vec<usize>,
    rank: Vec<usize>,
    size: Vec<usize>,
}

#[allow(dead_code)]
impl UnionFind {
    fn new(n: usize) -> UnionFind {
        let mut p = vec![0; n];
        for i in 0..n {
            p[i] = i;
        }
        return UnionFind {
            parent: p,
            rank: vec![0; n],
            size: vec![1; n],
        };
    }

    fn find(&mut self, x: usize) -> usize {
        if x == self.parent[x] {
            x
        } else {
            let p = self.parent[x];
            let pr = self.find(p);
            self.parent[x] = pr;
            pr
        }
    }

    fn same(&mut self, a: usize, b: usize) -> bool {
        self.find(a) == self.find(b)
    }

    fn unite(&mut self, a: usize, b: usize) {
        let a_root = self.find(a);
        let b_root = self.find(b);
        if self.rank[a_root] > self.rank[b_root] {
            self.parent[b_root] = a_root;
            self.size[a_root] += self.size[b_root];
        } else {
            self.parent[a_root] = b_root;
            self.size[b_root] += self.size[a_root];
            if self.rank[a_root] == self.rank[b_root] {
                self.rank[b_root] += 1;
            }
        }
    }

    fn get_size(&mut self, x: usize) -> usize {
        let root = self.find(x);
        self.size[root]
    }
}

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

    let mut g = vec![Vec::new(); n];
    let mut dist = vec![vec![std::usize::MAX; n]; 3];

    for (u, v) in uv.into_iter() {
        g[u].push(v);
    }

    let mut q = std::collections::VecDeque::new();
    q.push_front((0, s * 3));
    dist[s][0] = 0;

    while let Some((d, p)) = q.pop_front() {
        if p == t * 3 && d % 3 == 0 {
            return;
        }

        for &v in g[p / 3].iter() {
            if d + 1 < dist[(p + 1) % 3][v] {
                q.push_back((d + 1, v * 3 + (p + 1) % 3));
                dist[(p + 1) % 3][v] = d + 1;
            }
        }
    }

    println!("-1");
}