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::io::stdin;
use std::process;

// for i in &v[0] { for j in &v[*i as usize] { println!("0}", j ) } }

fn kenken(s: usize, ts: &Vec<Vec<usize>>, d: &mut Vec<usize>, count: usize, dest: &mut Vec<usize>) {
    for t1 in &ts[s] {
        for t2 in &ts[*t1] {
            for t3 in &ts[*t2] {
                if d[*t3] == 0 {
                    d[*t3] = count;
                    dest.push(*t3);
                }
            }
        }
    }
}

fn main() {
    let mut ss = String::new();
    stdin().read_line(&mut ss).ok();
    let n: usize;
    let m: usize;
    let s: usize;
    let t: usize;
    let v: Vec<usize> = ss
        .trim()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect();
    n = v[0];
    m = v[1];
    let mut uv: Vec<Vec<usize>> = Vec::new();
    for _i in 0..m {
        ss = String::from("");
        stdin().read_line(&mut ss).ok();
        uv.push(
            ss.trim()
                .split_whitespace()
                .map(|e| e.parse().ok().unwrap())
                .collect(),
        );
    }
    ss = String::from("");
    stdin().read_line(&mut ss).ok();
    let v: Vec<usize> = ss
        .trim()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect();
    s = v[0];
    t = v[1];
    let mut d: Vec<usize> = Vec::new();
    let mut ts: Vec<Vec<usize>> = Vec::new();
    for _i in 0..(n + 1) {
        d.push(0);
        ts.push(vec![]);
    }
    for e in uv {
        let u = e[0];
        let v = e[1];
        ts[u].push(v);
    }
    let mut count = 1;
    let mut dest: Vec<usize> = vec![s];
    loop {
        if dest.len() == 0 {
            break;
        }

        let mut ndest: Vec<usize> = Vec::new();
        for ss in dest {
            kenken(ss, &ts, &mut d, count, &mut ndest);
        }
        if d[t] > 0 {
            println!("{}", d[t]);
            process::exit(0);
        }
        count += 1;
        dest = ndest;
    }
    println!("-1");
}