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::vec::Vec;
use std::cmp::*;
use::std::collections::VecDeque;


fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
    let (n, m) = {
      let v: Vec<usize> = read_vec();
      (v[0], v[1])
    };

    let mut g: Vec<Vec<Vec<(usize, usize)>>> = vec![vec![vec![]; 3]; n];
    for _ in 0..m {
      let v: Vec<usize> = read_vec();
      g[v[0] - 1][0].push((v[1] - 1, 1));
      g[v[0] - 1][1].push((v[1] - 1, 2));
      g[v[0] - 1][2].push((v[1] - 1, 0));
    }
    let (s, t) = {
      let v: Vec<usize> = read_vec();
      (v[0] - 1, v[1] - 1)
    };

    match bfs_dist(n, &g, s, t) {
      std::u64::MAX => {println!("-1");},
      n => {println!("{}", n/3);},
    };


}

//単一視点から目的のノードまでの最短距離、ただし全ての辺の重みが等しいグラフに限る
fn bfs_dist(n: usize, g: &Vec<Vec<Vec<(usize, usize)>>>, start: usize, goal: usize) -> u64 {
  let mut visit = vec![vec![false; 3]; n];
  let mut q: VecDeque<(usize, usize)> = VecDeque::new();
  let mut dist = vec![vec![std::u64::MAX; 3]; n];

  //init
  visit[start][0] = true;
  q.push_back((start, 0));
  dist[start][0] = 0;

  while !q.is_empty() {
    let now
     = match q.pop_front() {
        None => (0, 0),
        Some(e) => e,
     };

    if goal == now.0 && now.1 == 0 {
      break;
    }

    for &e in &g[now.0][now.1] {
      if !visit[e.0][e.1] {
          visit[e.0][e.1] = true;
          dist[e.0][e.1] = min(dist[e.0][e.1], dist[now.0][now.1] + 1);
          q.push_back(e);
      }
    }
  }
  return dist[goal][0];
}