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)

use std::io;
use std::collections::VecDeque;

fn main() {
    let (n, m) = {
        let i = read::<usize>();
        (i[0], i[1])
    };
    let mut g = vec![Vec::new(); n];
    for _ in 0..m {
        let (u, v) = {
            let i = read::<usize>();
            (i[0] - 1, i[1] - 1)
        };
        g[u].push(v);
    }
    let (s, t) = {
        let i = read::<usize>();
        (i[0] - 1, i[1] - 1)
    };

    let mut que = VecDeque::new();
    let mut dist = vec![vec![::std::usize::MAX; 3]; n];
    que.push_back((s, 0));
    while let Some((u, v)) = que.pop_front() {
        if dist[u][v % 3] != ::std::usize::MAX { continue; }
        dist[u][v % 3] = v;

        for i in 0..g[u].len() {
            let vv = g[u][i];
            que.push_back((vv, v + 1));
        }
    }

    if dist[t][0] != ::std::usize::MAX { println!("{}", dist[t][0] / 3); }
    else { println!("-1"); }
}

#[allow(dead_code)]
fn read<T>() -> Vec<T>
where
    T: std::str::FromStr,
    T::Err: std::fmt::Debug,
{
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    buf.split_whitespace()
        .map(|s| s.trim().parse().unwrap())
        .collect()
}

#[allow(dead_code)]
fn read_one<T>() -> T
where
    T: std::str::FromStr,
    T::Err: std::fmt::Debug,
{
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    buf.trim().parse().unwrap()
}