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::collections::VecDeque;

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

    let mut stack = VecDeque::new();
    let mut dist = vec![-3; 3*n];

    stack.push_back((s,0));
    dist[3*s] = 0;
    while let Some((v, a)) = stack.pop_front() {
        let b = (a+1)%3;
        for &w in &neighbors[v] {
            if dist[3*w+b] == -3
                || dist[3*w+b] > dist[3*v+a] + 1 {
                dist[3*w+b] = dist[3*v+a] + 1;
                stack.push_back((w,b));
            }
        }
    }

    println!("{}", dist[3*t] / 3);
}

fn get_line() -> String {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).ok();
    buf.trim().to_owned()
}
fn read<T: std::str::FromStr>() -> T{
    get_line().parse().ok().unwrap()
}
fn read_vec<T: std::str::FromStr>() -> Vec<T>{
    get_line().split_whitespace()
        .map(|e| e.parse().ok().unwrap()).collect()
}
fn read_pair<T: std::str::FromStr + Copy>() -> (T,T){
    let v:Vec<T> = read_vec(); (v[0], v[1])
}