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::*;

fn main(){
    let n : usize = input();
    let m : usize = input();
    let mut e : Vec<Vec<usize>> = vec![vec![]; n * 3];
    for _ in 0..m{
        let mut u : usize = input();
        let mut v : usize = input();
        u -= 1;
        v -= 1;
        e[u].push(v + n);
        e[u + n].push(v + n * 2);
        e[u + n * 2].push(v);
    }
    let mut s : usize = input();
    let mut t : usize = input();
    s -= 1;
    t -= 1;
    let mut d : Vec<isize> = vec![std::isize::MAX; n * 3];
    d[s] = 0;
    dfs(s, &e, &mut d, &t);
    let ans = if d[t] != std::isize::MAX {d[t] / 3}else{-1};
    println!("{}", ans);
}

fn dfs(p:usize, e:&Vec<Vec<usize>>, d:&mut Vec<isize>, t: &usize){
    for i in 0..(e[p].len()){
        let to = e[p][i];
        // println!("{} {}", d[to], d[p]);
        if d[to] > d[p] + 1{
            d[to] = d[p] + 1;
            if to == *t{
                break;
            }
            dfs(to, &e, d, &t);
        }
    }
}

fn input<T>() -> T
where
    T: std::str::FromStr,
{
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}