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)

use std::collections::HashSet;
use std::io::{stdin, BufRead};


fn make_3hop(g: &[Vec<usize>], start: usize) -> HashSet<usize> {
    let mut nei = HashSet::new();
    nei.insert(start);
    for _ in 0..3 {
        let mut new = HashSet::with_capacity(8);
        for &u in nei.iter() {
            for v in g[u].iter() {
                new.insert(*v);
            }
        }
        nei = new;
    }
    nei
}

fn bfs(g: &[Vec<usize>], start: usize, goal: usize) -> i32 {
    let mut gg = Vec::with_capacity(g.len());
    for _ in 0..g.len() {
        gg.push((true, HashSet::new()));
    }
    let mut visited = vec![false; gg.len()];
    let mut now = vec![start];
    visited[start] = true;
    let mut step = 1;
    while !now.is_empty() {
        let mut nex = Vec::new();
        for u in now {
            if gg[u].0 {
                gg[u] = (false, make_3hop(g, u));
            }
            for &v in gg[u].1.iter() {
                if visited[v] {
                    continue;
                }
                if v == goal {
                    return step;
                }
                visited[v] = true;
                nex.push(v);
            }
        }
        now = nex;
        step += 1;
    }
    -1
}

fn main() {
    let stdin = stdin();
    let cin = stdin.lock();
    let mut lines = cin.lines();
    let (n, m) = {
        let line = lines.next().unwrap().unwrap();
        let mut words = line.split(' ');
        (
            words.next().unwrap().parse::<usize>().unwrap(),
            words.next().unwrap().parse::<usize>().unwrap(),
        )
    };

    let mut g = Vec::with_capacity(n);
    for _ in 0..n {
        g.push(Vec::new());
    }
    for _ in 0..m {
        let (u, v) = {
            let line = lines.next().unwrap().unwrap();
            let mut words = line.split(' ');
            (
                words.next().unwrap().parse::<usize>().unwrap() - 1,
                words.next().unwrap().parse::<usize>().unwrap() - 1,
            )
        };
        g[u].push(v);
    }
    let (s, t) = {
        let line = lines.next().unwrap().unwrap();
        let mut words = line.split(' ');
        (
            words.next().unwrap().parse::<usize>().unwrap() - 1,
            words.next().unwrap().parse::<usize>().unwrap() - 1,
        )
    };
    println!("{}", bfs(&g, s, t));
}